Wordpress ACF get_field not loading

I've been racking my brain on this one. I've looked through the related stack overflow questions and yet can not seem to piece together this.

I am attempting to render a link using custom fields in WordPress and the get_field() function. I'm a new developer so I'm attempting to just console log it yet I keep returning "empty data" and "null" for my results. I understand get_field() will return 2 results, but I can't seem to extract my desired links. Any help is appreciated.

Custom fields:

image

This is my code base:

    $trend_one = get_field('trend_one');

    $trend_two = get_field('trend_two');

    $trend_three = get_field('trend_three');





?>

<div class='trend_bar'>

    <div class="trend_item">TRENDING:</div>

    <div class='trend_item'><?php console_log($trend_one); ?></div>

    <div class='trend_item'><?php console_log($trend_two); ?></div>

    <div class='trend_item'><?php console_log($trend_three); ?></div>

</div>

edited Nov 21 '18 at 15:30

Gufran Hasan

3,57241426

asked Nov 21 '18 at 15:28

Marsel Gray

147

console.log is a javascript function. Not sure where you picked up console_log - but if it's a function you've written / added, you need to share that code. There's no need, really, though - In PHP, you'd use var_dump($trend_one) (or var_export). Note you may get recommendations to use echo or print_r, but those are limited - echo won't work for objects / arrays, and print_r will only work for objects / arrays, whereas var_dump or var_export will handle any variable contents.

– cale_b
Nov 21 '18 at 15:47

Please read how to create an Minimal, Complete, and Verifiable example - while your code is part of the equation, there's much more to the story, and if we don't know where you've put this code specifically, that could be the problem. get_field will only work in the right context (when the WP post has been set up), so knowing where in your template, which template, etc. matters.

– cale_b
Nov 21 '18 at 15:52

There are cases when the field name won’t work - in these cases you have to use field ID (that’s a long, unique code for the field, the ACF documentation tells you how to get it). And the other thing: try to set the second parameter of the get_field() function - that’s the post ID of the WP object whose ACF field you’re trying to get.

– muka.gergely
Nov 21 '18 at 18:24

add a comment |

I've been racking my brain on this one. I've looked through the related stack overflow questions and yet can not seem to piece together this.

Custom fields:

image

This is my code base:

    $trend_one = get_field('trend_one');

    $trend_two = get_field('trend_two');

    $trend_three = get_field('trend_three');





?>

<div class='trend_bar'>

    <div class="trend_item">TRENDING:</div>

    <div class='trend_item'><?php console_log($trend_one); ?></div>

    <div class='trend_item'><?php console_log($trend_two); ?></div>

    <div class='trend_item'><?php console_log($trend_three); ?></div>

</div>

edited Nov 21 '18 at 15:30

Gufran Hasan

3,57241426

asked Nov 21 '18 at 15:28

Marsel Gray

147

console.log is a javascript function. Not sure where you picked up console_log - but if it's a function you've written / added, you need to share that code. There's no need, really, though - In PHP, you'd use var_dump($trend_one) (or var_export). Note you may get recommendations to use echo or print_r, but those are limited - echo won't work for objects / arrays, and print_r will only work for objects / arrays, whereas var_dump or var_export will handle any variable contents.

– cale_b
Nov 21 '18 at 15:47

Please read how to create an Minimal, Complete, and Verifiable example - while your code is part of the equation, there's much more to the story, and if we don't know where you've put this code specifically, that could be the problem. get_field will only work in the right context (when the WP post has been set up), so knowing where in your template, which template, etc. matters.

– cale_b
Nov 21 '18 at 15:52

There are cases when the field name won’t work - in these cases you have to use field ID (that’s a long, unique code for the field, the ACF documentation tells you how to get it). And the other thing: try to set the second parameter of the get_field() function - that’s the post ID of the WP object whose ACF field you’re trying to get.

– muka.gergely
Nov 21 '18 at 18:24

add a comment |

I've been racking my brain on this one. I've looked through the related stack overflow questions and yet can not seem to piece together this.

Custom fields:

image

This is my code base:

    $trend_one = get_field('trend_one');

    $trend_two = get_field('trend_two');

    $trend_three = get_field('trend_three');





?>

<div class='trend_bar'>

    <div class="trend_item">TRENDING:</div>

    <div class='trend_item'><?php console_log($trend_one); ?></div>

    <div class='trend_item'><?php console_log($trend_two); ?></div>

    <div class='trend_item'><?php console_log($trend_three); ?></div>

</div>

edited Nov 21 '18 at 15:30

Gufran Hasan

3,57241426

asked Nov 21 '18 at 15:28

Marsel Gray

147

I've been racking my brain on this one. I've looked through the related stack overflow questions and yet can not seem to piece together this.

Custom fields:

image

This is my code base:

    $trend_one = get_field('trend_one');

    $trend_two = get_field('trend_two');

    $trend_three = get_field('trend_three');





?>

<div class='trend_bar'>

    <div class="trend_item">TRENDING:</div>

    <div class='trend_item'><?php console_log($trend_one); ?></div>

    <div class='trend_item'><?php console_log($trend_two); ?></div>

    <div class='trend_item'><?php console_log($trend_three); ?></div>

</div>

php wordpress advanced-custom-fields

edited Nov 21 '18 at 15:30

Gufran Hasan

3,57241426

asked Nov 21 '18 at 15:28

Marsel Gray

147

edited Nov 21 '18 at 15:30

Gufran Hasan

3,57241426

asked Nov 21 '18 at 15:28

Marsel Gray

147

edited Nov 21 '18 at 15:30

Gufran Hasan

3,57241426

edited Nov 21 '18 at 15:30

Gufran Hasan

3,57241426

edited Nov 21 '18 at 15:30

Gufran Hasan

3,57241426

asked Nov 21 '18 at 15:28

Marsel Gray

147

asked Nov 21 '18 at 15:28

Marsel Gray

147

asked Nov 21 '18 at 15:28

Marsel Gray

147

console.log is a javascript function. Not sure where you picked up console_log - but if it's a function you've written / added, you need to share that code. There's no need, really, though - In PHP, you'd use var_dump($trend_one) (or var_export). Note you may get recommendations to use echo or print_r, but those are limited - echo won't work for objects / arrays, and print_r will only work for objects / arrays, whereas var_dump or var_export will handle any variable contents.

– cale_b
Nov 21 '18 at 15:47

Please read how to create an Minimal, Complete, and Verifiable example - while your code is part of the equation, there's much more to the story, and if we don't know where you've put this code specifically, that could be the problem. get_field will only work in the right context (when the WP post has been set up), so knowing where in your template, which template, etc. matters.

– cale_b
Nov 21 '18 at 15:52

There are cases when the field name won’t work - in these cases you have to use field ID (that’s a long, unique code for the field, the ACF documentation tells you how to get it). And the other thing: try to set the second parameter of the get_field() function - that’s the post ID of the WP object whose ACF field you’re trying to get.

– muka.gergely
Nov 21 '18 at 18:24

add a comment |

console.log is a javascript function. Not sure where you picked up console_log - but if it's a function you've written / added, you need to share that code. There's no need, really, though - In PHP, you'd use var_dump($trend_one) (or var_export). Note you may get recommendations to use echo or print_r, but those are limited - echo won't work for objects / arrays, and print_r will only work for objects / arrays, whereas var_dump or var_export will handle any variable contents.

– cale_b
Nov 21 '18 at 15:47

Please read how to create an Minimal, Complete, and Verifiable example - while your code is part of the equation, there's much more to the story, and if we don't know where you've put this code specifically, that could be the problem. get_field will only work in the right context (when the WP post has been set up), so knowing where in your template, which template, etc. matters.

– cale_b
Nov 21 '18 at 15:52

There are cases when the field name won’t work - in these cases you have to use field ID (that’s a long, unique code for the field, the ACF documentation tells you how to get it). And the other thing: try to set the second parameter of the get_field() function - that’s the post ID of the WP object whose ACF field you’re trying to get.

– muka.gergely
Nov 21 '18 at 18:24

console.log is a javascript function. Not sure where you picked up console_log - but if it's a function you've written / added, you need to share that code. There's no need, really, though - In PHP, you'd use var_dump($trend_one) (or var_export). Note you may get recommendations to use echo or print_r, but those are limited - echo won't work for objects / arrays, and print_r will only work for objects / arrays, whereas var_dump or var_export will handle any variable contents.

– cale_b
Nov 21 '18 at 15:47

Please read how to create an Minimal, Complete, and Verifiable example - while your code is part of the equation, there's much more to the story, and if we don't know where you've put this code specifically, that could be the problem. get_field will only work in the right context (when the WP post has been set up), so knowing where in your template, which template, etc. matters.

– cale_b
Nov 21 '18 at 15:52

There are cases when the field name won’t work - in these cases you have to use field ID (that’s a long, unique code for the field, the ACF documentation tells you how to get it). And the other thing: try to set the second parameter of the get_field() function - that’s the post ID of the WP object whose ACF field you’re trying to get.

– muka.gergely
Nov 21 '18 at 18:24

add a comment |

2 Answers
2

active

oldest

votes

Why are you using console.log inside PHP?

It should be:

<?php echo $trend_one['url']; ?>

<?php echo $trend_two; ?>

<?php echo $trend_three; ?>

https://www.advancedcustomfields.com/resources/link/

answered Nov 21 '18 at 15:45

tadadadadi

135

Thanks for wanting to help! Note that echo won't handle some variable types (arrays, objects, booleans) - may I suggest that you propose var_dump or var_export instead?

– cale_b
Nov 21 '18 at 15:48

@cale_b Hi, thanks. I wasn't sure if the OP is asking about the correct way to output the value, or the correct way to debug. Since console.log is not PHP, I thought the answer might be as simple as using echo. Shouldn't we get a PHP error, if we use console.log? - which is why I was a bit confused, how the code returns "null" or "empty data".

– tadadadadi
Nov 21 '18 at 16:16

Agreed on all points. My only point is that echo array(1,2,3); will display the word Array on the screen, and you might want to suggest var_dump instead :)

– cale_b
Nov 21 '18 at 16:19

add a comment |

Here's that code you wanted for the console_log in PHP

function console_log( $data ) {

if ( empty( $data ) ){

    print '<script>console.log("Empty Data");</script>';

    print '<script>console.log(' . json_encode( $data ) . ');</script>';

} else {

    if ( is_array( $data ) || is_object( $data ) ){

        print '<script>console.table(' . json_encode( $data ) . ');</script>';

        print '<script>console.log(' . json_encode( $data ) . ');</script>';



    } else {

        print '<script>console.log(' . json_encode( $data ) . ');</script>';

    }

}

}

answered Nov 21 '18 at 15:59

Marsel Gray

147

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Why are you using console.log inside PHP?

It should be:

<?php echo $trend_one['url']; ?>

<?php echo $trend_two; ?>

<?php echo $trend_three; ?>

https://www.advancedcustomfields.com/resources/link/

answered Nov 21 '18 at 15:45

tadadadadi

135

Thanks for wanting to help! Note that echo won't handle some variable types (arrays, objects, booleans) - may I suggest that you propose var_dump or var_export instead?

– cale_b
Nov 21 '18 at 15:48

@cale_b Hi, thanks. I wasn't sure if the OP is asking about the correct way to output the value, or the correct way to debug. Since console.log is not PHP, I thought the answer might be as simple as using echo. Shouldn't we get a PHP error, if we use console.log? - which is why I was a bit confused, how the code returns "null" or "empty data".

– tadadadadi
Nov 21 '18 at 16:16

Agreed on all points. My only point is that echo array(1,2,3); will display the word Array on the screen, and you might want to suggest var_dump instead :)

– cale_b
Nov 21 '18 at 16:19

add a comment |

Why are you using console.log inside PHP?

It should be:

<?php echo $trend_one['url']; ?>

<?php echo $trend_two; ?>

<?php echo $trend_three; ?>

https://www.advancedcustomfields.com/resources/link/

answered Nov 21 '18 at 15:45

tadadadadi

135

Thanks for wanting to help! Note that echo won't handle some variable types (arrays, objects, booleans) - may I suggest that you propose var_dump or var_export instead?

– cale_b
Nov 21 '18 at 15:48

@cale_b Hi, thanks. I wasn't sure if the OP is asking about the correct way to output the value, or the correct way to debug. Since console.log is not PHP, I thought the answer might be as simple as using echo. Shouldn't we get a PHP error, if we use console.log? - which is why I was a bit confused, how the code returns "null" or "empty data".

– tadadadadi
Nov 21 '18 at 16:16

Agreed on all points. My only point is that echo array(1,2,3); will display the word Array on the screen, and you might want to suggest var_dump instead :)

– cale_b
Nov 21 '18 at 16:19

add a comment |

Why are you using console.log inside PHP?

It should be:

<?php echo $trend_one['url']; ?>

<?php echo $trend_two; ?>

<?php echo $trend_three; ?>

https://www.advancedcustomfields.com/resources/link/

answered Nov 21 '18 at 15:45

tadadadadi

135

Why are you using console.log inside PHP?

It should be:

<?php echo $trend_one['url']; ?>

<?php echo $trend_two; ?>

<?php echo $trend_three; ?>

https://www.advancedcustomfields.com/resources/link/

answered Nov 21 '18 at 15:45

tadadadadi

135

answered Nov 21 '18 at 15:45

tadadadadi

135

answered Nov 21 '18 at 15:45

tadadadadi

135

answered Nov 21 '18 at 15:45

tadadadadi

135

Thanks for wanting to help! Note that echo won't handle some variable types (arrays, objects, booleans) - may I suggest that you propose var_dump or var_export instead?

– cale_b
Nov 21 '18 at 15:48

@cale_b Hi, thanks. I wasn't sure if the OP is asking about the correct way to output the value, or the correct way to debug. Since console.log is not PHP, I thought the answer might be as simple as using echo. Shouldn't we get a PHP error, if we use console.log? - which is why I was a bit confused, how the code returns "null" or "empty data".

– tadadadadi
Nov 21 '18 at 16:16

Agreed on all points. My only point is that echo array(1,2,3); will display the word Array on the screen, and you might want to suggest var_dump instead :)

– cale_b
Nov 21 '18 at 16:19

add a comment |

Thanks for wanting to help! Note that echo won't handle some variable types (arrays, objects, booleans) - may I suggest that you propose var_dump or var_export instead?

– cale_b
Nov 21 '18 at 15:48

@cale_b Hi, thanks. I wasn't sure if the OP is asking about the correct way to output the value, or the correct way to debug. Since console.log is not PHP, I thought the answer might be as simple as using echo. Shouldn't we get a PHP error, if we use console.log? - which is why I was a bit confused, how the code returns "null" or "empty data".

– tadadadadi
Nov 21 '18 at 16:16

Agreed on all points. My only point is that echo array(1,2,3); will display the word Array on the screen, and you might want to suggest var_dump instead :)

– cale_b
Nov 21 '18 at 16:19

Thanks for wanting to help! Note that echo won't handle some variable types (arrays, objects, booleans) - may I suggest that you propose var_dump or var_export instead?

– cale_b
Nov 21 '18 at 15:48

@cale_b Hi, thanks. I wasn't sure if the OP is asking about the correct way to output the value, or the correct way to debug. Since console.log is not PHP, I thought the answer might be as simple as using echo. Shouldn't we get a PHP error, if we use console.log? - which is why I was a bit confused, how the code returns "null" or "empty data".

– tadadadadi
Nov 21 '18 at 16:16

Agreed on all points. My only point is that echo array(1,2,3); will display the word Array on the screen, and you might want to suggest var_dump instead :)

– cale_b
Nov 21 '18 at 16:19

add a comment |

Here's that code you wanted for the console_log in PHP

function console_log( $data ) {

if ( empty( $data ) ){

    print '<script>console.log("Empty Data");</script>';

    print '<script>console.log(' . json_encode( $data ) . ');</script>';

} else {

    if ( is_array( $data ) || is_object( $data ) ){

        print '<script>console.table(' . json_encode( $data ) . ');</script>';

        print '<script>console.log(' . json_encode( $data ) . ');</script>';



    } else {

        print '<script>console.log(' . json_encode( $data ) . ');</script>';

    }

}

}

answered Nov 21 '18 at 15:59

Marsel Gray

147

add a comment |

Here's that code you wanted for the console_log in PHP

function console_log( $data ) {

if ( empty( $data ) ){

    print '<script>console.log("Empty Data");</script>';

    print '<script>console.log(' . json_encode( $data ) . ');</script>';

} else {

    if ( is_array( $data ) || is_object( $data ) ){

        print '<script>console.table(' . json_encode( $data ) . ');</script>';

        print '<script>console.log(' . json_encode( $data ) . ');</script>';



    } else {

        print '<script>console.log(' . json_encode( $data ) . ');</script>';

    }

}

}

answered Nov 21 '18 at 15:59

Marsel Gray

147

add a comment |

Here's that code you wanted for the console_log in PHP

function console_log( $data ) {

if ( empty( $data ) ){

    print '<script>console.log("Empty Data");</script>';

    print '<script>console.log(' . json_encode( $data ) . ');</script>';

} else {

    if ( is_array( $data ) || is_object( $data ) ){

        print '<script>console.table(' . json_encode( $data ) . ');</script>';

        print '<script>console.log(' . json_encode( $data ) . ');</script>';



    } else {

        print '<script>console.log(' . json_encode( $data ) . ');</script>';

    }

}

}

answered Nov 21 '18 at 15:59

Marsel Gray

147

Here's that code you wanted for the console_log in PHP

function console_log( $data ) {

if ( empty( $data ) ){

    print '<script>console.log("Empty Data");</script>';

    print '<script>console.log(' . json_encode( $data ) . ');</script>';

} else {

    if ( is_array( $data ) || is_object( $data ) ){

        print '<script>console.table(' . json_encode( $data ) . ');</script>';

        print '<script>console.log(' . json_encode( $data ) . ');</script>';



    } else {

        print '<script>console.log(' . json_encode( $data ) . ');</script>';

    }

}

}

answered Nov 21 '18 at 15:59

Marsel Gray

147

answered Nov 21 '18 at 15:59

Marsel Gray

147

answered Nov 21 '18 at 15:59

Marsel Gray

147

answered Nov 21 '18 at 15:59

Marsel Gray

147

add a comment |

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