Checking if 2 php variables are set inside Javascript isset

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I have 2 php variables in PHP (mainly $usm and $ag) and am passing them to the frontend. In Javascript am using isset to check if they have a value before executing some code but seems not to work

<script>

    if( <?php isset($usm , $ag) ?> ){

        $( document ).ready(function() {

            var usmData = {!! json_encode($usm) !!};

            var agData = {!! json_encode($ag) !!};

        });

    }

</script

asked Jan 3 at 9:52

Martin

1841111

You may not access the PHP variable within the script. Try to set the php variable in cookies or sessions and extract the php variable data from there.

– Ramya
Jan 3 at 9:55

add a comment |

I have 2 php variables in PHP (mainly $usm and $ag) and am passing them to the frontend. In Javascript am using isset to check if they have a value before executing some code but seems not to work

<script>

    if( <?php isset($usm , $ag) ?> ){

        $( document ).ready(function() {

            var usmData = {!! json_encode($usm) !!};

            var agData = {!! json_encode($ag) !!};

        });

    }

</script

asked Jan 3 at 9:52

Martin

1841111

You may not access the PHP variable within the script. Try to set the php variable in cookies or sessions and extract the php variable data from there.

– Ramya
Jan 3 at 9:55

add a comment |

I have 2 php variables in PHP (mainly $usm and $ag) and am passing them to the frontend. In Javascript am using isset to check if they have a value before executing some code but seems not to work

<script>

    if( <?php isset($usm , $ag) ?> ){

        $( document ).ready(function() {

            var usmData = {!! json_encode($usm) !!};

            var agData = {!! json_encode($ag) !!};

        });

    }

</script

asked Jan 3 at 9:52

Martin

1841111

I have 2 php variables in PHP (mainly $usm and $ag) and am passing them to the frontend. In Javascript am using isset to check if they have a value before executing some code but seems not to work

<script>

    if( <?php isset($usm , $ag) ?> ){

        $( document ).ready(function() {

            var usmData = {!! json_encode($usm) !!};

            var agData = {!! json_encode($ag) !!};

        });

    }

</script

javascript php variables

asked Jan 3 at 9:52

Martin

1841111

asked Jan 3 at 9:52

Martin

1841111

asked Jan 3 at 9:52

Martin

1841111

asked Jan 3 at 9:52

Martin

1841111

asked Jan 3 at 9:52

Martin

1841111

You may not access the PHP variable within the script. Try to set the php variable in cookies or sessions and extract the php variable data from there.

– Ramya
Jan 3 at 9:55

add a comment |

You may not access the PHP variable within the script. Try to set the php variable in cookies or sessions and extract the php variable data from there.

– Ramya
Jan 3 at 9:55

You may not access the PHP variable within the script. Try to set the php variable in cookies or sessions and extract the php variable data from there.

– Ramya
Jan 3 at 9:55

add a comment |

4 Answers
4

active

oldest

votes

You need to check the isset part with PHP only.

<?php if(isset($usm) and isset($ag)){ ?>

<script>

    $( document ).ready(function() {

        var usmData = '<?php echo json_encode($usm); ?>';

        var agData = '<?php echo json_encode($ag); ?>';

    });

</script>

<?php } ?>

edited Jan 3 at 10:14

answered Jan 3 at 9:59

executable

1,9832924

JS strings need to be surrounded by ' or "

– Vahid Amiri
Jan 3 at 10:05

Yes fixed also his encoding in JSON

– executable
Jan 3 at 10:06

@executable I tried your solution and I get this error Parse error: syntax error, unexpected '{'

– Martin
Jan 3 at 10:08

Can you show a sample data of $usm and $ag

– executable
Jan 3 at 10:08

Noted the error,, it works fine thanks alot..

– Martin
Jan 3 at 10:13

add a comment |

That doesn't make sense. You mixed up PHP and JS.

Result of PHP isset is only used on PHP and doesn't make it to JS side. Your JS if statement is not valid.

Check this article for more info.

edited Jan 3 at 11:12

MorganFreeFarm

1,48431328

answered Jan 3 at 9:57

TuckerS

3313

But I placed it inside php tags <?php ... ?>

– Martin
Jan 3 at 9:58

1

@Martin It doesn't matter because Javascript expects a boolean value inside if and your the result of your PHP code doesn't get passed to JS. not like that anyway.

– TuckerS
Jan 3 at 10:01

add a comment |

<script>

    <?php if(isset($usm , $ag)) { ?> 

        $( document ).ready(function() {

            var usmData = {!! json_encode($usm) !!};

            var agData = {!! json_encode($ag) !!};

        });



    <?php } ?>

</script>

answered Jan 3 at 10:00

TDK

961720

add a comment |

I would suggest you to write the PHP code outside otherwise script tag will be executed unnecessary with empty code if the condition is not met.

<?php 

    if(isset($usm , $ag) {

?>

<script>

    $( document ).ready(function() {

        var usmData = '<?php echo json_encode($usm); ?>';

        var agData = '<?php echo json_encode($ag); ?>';

    });

</script>

<?php

    }

?>

answered Jan 3 at 10:11

Rajendra arora

1,51211018

add a comment |

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

You need to check the isset part with PHP only.

<?php if(isset($usm) and isset($ag)){ ?>

<script>

    $( document ).ready(function() {

        var usmData = '<?php echo json_encode($usm); ?>';

        var agData = '<?php echo json_encode($ag); ?>';

    });

</script>

<?php } ?>

edited Jan 3 at 10:14

answered Jan 3 at 9:59

executable

1,9832924

JS strings need to be surrounded by ' or "

– Vahid Amiri
Jan 3 at 10:05

Yes fixed also his encoding in JSON

– executable
Jan 3 at 10:06

@executable I tried your solution and I get this error Parse error: syntax error, unexpected '{'

– Martin
Jan 3 at 10:08

Can you show a sample data of $usm and $ag

– executable
Jan 3 at 10:08

Noted the error,, it works fine thanks alot..

– Martin
Jan 3 at 10:13

add a comment |

You need to check the isset part with PHP only.

<?php if(isset($usm) and isset($ag)){ ?>

<script>

    $( document ).ready(function() {

        var usmData = '<?php echo json_encode($usm); ?>';

        var agData = '<?php echo json_encode($ag); ?>';

    });

</script>

<?php } ?>

edited Jan 3 at 10:14

answered Jan 3 at 9:59

executable

1,9832924

JS strings need to be surrounded by ' or "

– Vahid Amiri
Jan 3 at 10:05

Yes fixed also his encoding in JSON

– executable
Jan 3 at 10:06

@executable I tried your solution and I get this error Parse error: syntax error, unexpected '{'

– Martin
Jan 3 at 10:08

Can you show a sample data of $usm and $ag

– executable
Jan 3 at 10:08

Noted the error,, it works fine thanks alot..

– Martin
Jan 3 at 10:13

add a comment |

You need to check the isset part with PHP only.

<?php if(isset($usm) and isset($ag)){ ?>

<script>

    $( document ).ready(function() {

        var usmData = '<?php echo json_encode($usm); ?>';

        var agData = '<?php echo json_encode($ag); ?>';

    });

</script>

<?php } ?>

edited Jan 3 at 10:14

answered Jan 3 at 9:59

executable

1,9832924

You need to check the isset part with PHP only.

<?php if(isset($usm) and isset($ag)){ ?>

<script>

    $( document ).ready(function() {

        var usmData = '<?php echo json_encode($usm); ?>';

        var agData = '<?php echo json_encode($ag); ?>';

    });

</script>

<?php } ?>

edited Jan 3 at 10:14

answered Jan 3 at 9:59

executable

1,9832924

edited Jan 3 at 10:14

answered Jan 3 at 9:59

executable

1,9832924

answered Jan 3 at 9:59

executable

1,9832924

answered Jan 3 at 9:59

executable

1,9832924

JS strings need to be surrounded by ' or "

– Vahid Amiri
Jan 3 at 10:05

Yes fixed also his encoding in JSON

– executable
Jan 3 at 10:06

@executable I tried your solution and I get this error Parse error: syntax error, unexpected '{'

– Martin
Jan 3 at 10:08

Can you show a sample data of $usm and $ag

– executable
Jan 3 at 10:08

Noted the error,, it works fine thanks alot..

– Martin
Jan 3 at 10:13

add a comment |

JS strings need to be surrounded by ' or "

– Vahid Amiri
Jan 3 at 10:05

Yes fixed also his encoding in JSON

– executable
Jan 3 at 10:06

@executable I tried your solution and I get this error Parse error: syntax error, unexpected '{'

– Martin
Jan 3 at 10:08

Can you show a sample data of $usm and $ag

– executable
Jan 3 at 10:08

Noted the error,, it works fine thanks alot..

– Martin
Jan 3 at 10:13

JS strings need to be surrounded by ' or "

– Vahid Amiri
Jan 3 at 10:05

Yes fixed also his encoding in JSON

– executable
Jan 3 at 10:06

@executable I tried your solution and I get this error Parse error: syntax error, unexpected '{'

– Martin
Jan 3 at 10:08

Can you show a sample data of $usm and $ag

– executable
Jan 3 at 10:08

Noted the error,, it works fine thanks alot..

– Martin
Jan 3 at 10:13

add a comment |

That doesn't make sense. You mixed up PHP and JS.

Result of PHP isset is only used on PHP and doesn't make it to JS side. Your JS if statement is not valid.

Check this article for more info.

edited Jan 3 at 11:12

MorganFreeFarm

1,48431328

answered Jan 3 at 9:57

TuckerS

3313

But I placed it inside php tags <?php ... ?>

– Martin
Jan 3 at 9:58

1

@Martin It doesn't matter because Javascript expects a boolean value inside if and your the result of your PHP code doesn't get passed to JS. not like that anyway.

– TuckerS
Jan 3 at 10:01

add a comment |

That doesn't make sense. You mixed up PHP and JS.

Result of PHP isset is only used on PHP and doesn't make it to JS side. Your JS if statement is not valid.

Check this article for more info.

edited Jan 3 at 11:12

MorganFreeFarm

1,48431328

answered Jan 3 at 9:57

TuckerS

3313

But I placed it inside php tags <?php ... ?>

– Martin
Jan 3 at 9:58

1

@Martin It doesn't matter because Javascript expects a boolean value inside if and your the result of your PHP code doesn't get passed to JS. not like that anyway.

– TuckerS
Jan 3 at 10:01

add a comment |

That doesn't make sense. You mixed up PHP and JS.

Result of PHP isset is only used on PHP and doesn't make it to JS side. Your JS if statement is not valid.

Check this article for more info.

edited Jan 3 at 11:12

MorganFreeFarm

1,48431328

answered Jan 3 at 9:57

TuckerS

3313

That doesn't make sense. You mixed up PHP and JS.

Result of PHP isset is only used on PHP and doesn't make it to JS side. Your JS if statement is not valid.

Check this article for more info.

edited Jan 3 at 11:12

MorganFreeFarm

1,48431328

answered Jan 3 at 9:57

TuckerS

3313

edited Jan 3 at 11:12

MorganFreeFarm

1,48431328

edited Jan 3 at 11:12

MorganFreeFarm

1,48431328

edited Jan 3 at 11:12

MorganFreeFarm

1,48431328

answered Jan 3 at 9:57

TuckerS

3313

answered Jan 3 at 9:57

TuckerS

3313

answered Jan 3 at 9:57

TuckerS

3313

But I placed it inside php tags <?php ... ?>

– Martin
Jan 3 at 9:58

1

@Martin It doesn't matter because Javascript expects a boolean value inside if and your the result of your PHP code doesn't get passed to JS. not like that anyway.

– TuckerS
Jan 3 at 10:01

add a comment |

But I placed it inside php tags <?php ... ?>

– Martin
Jan 3 at 9:58

1

@Martin It doesn't matter because Javascript expects a boolean value inside if and your the result of your PHP code doesn't get passed to JS. not like that anyway.

– TuckerS
Jan 3 at 10:01

But I placed it inside php tags <?php ... ?>

– Martin
Jan 3 at 9:58

@Martin It doesn't matter because Javascript expects a boolean value inside if and your the result of your PHP code doesn't get passed to JS. not like that anyway.

– TuckerS
Jan 3 at 10:01

add a comment |

<script>

    <?php if(isset($usm , $ag)) { ?> 

        $( document ).ready(function() {

            var usmData = {!! json_encode($usm) !!};

            var agData = {!! json_encode($ag) !!};

        });



    <?php } ?>

</script>

answered Jan 3 at 10:00

TDK

961720

add a comment |

<script>

    <?php if(isset($usm , $ag)) { ?> 

        $( document ).ready(function() {

            var usmData = {!! json_encode($usm) !!};

            var agData = {!! json_encode($ag) !!};

        });



    <?php } ?>

</script>

answered Jan 3 at 10:00

TDK

961720

add a comment |

<script>

    <?php if(isset($usm , $ag)) { ?> 

        $( document ).ready(function() {

            var usmData = {!! json_encode($usm) !!};

            var agData = {!! json_encode($ag) !!};

        });



    <?php } ?>

</script>

answered Jan 3 at 10:00

TDK

961720

<script>

    <?php if(isset($usm , $ag)) { ?> 

        $( document ).ready(function() {

            var usmData = {!! json_encode($usm) !!};

            var agData = {!! json_encode($ag) !!};

        });



    <?php } ?>

</script>

answered Jan 3 at 10:00

TDK

961720

answered Jan 3 at 10:00

TDK

961720

answered Jan 3 at 10:00

TDK

961720

answered Jan 3 at 10:00

TDK

961720

add a comment |

I would suggest you to write the PHP code outside otherwise script tag will be executed unnecessary with empty code if the condition is not met.

<?php 

    if(isset($usm , $ag) {

?>

<script>

    $( document ).ready(function() {

        var usmData = '<?php echo json_encode($usm); ?>';

        var agData = '<?php echo json_encode($ag); ?>';

    });

</script>

<?php

    }

?>

answered Jan 3 at 10:11

Rajendra arora

1,51211018

add a comment |

I would suggest you to write the PHP code outside otherwise script tag will be executed unnecessary with empty code if the condition is not met.

<?php 

    if(isset($usm , $ag) {

?>

<script>

    $( document ).ready(function() {

        var usmData = '<?php echo json_encode($usm); ?>';

        var agData = '<?php echo json_encode($ag); ?>';

    });

</script>

<?php

    }

?>

answered Jan 3 at 10:11

Rajendra arora

1,51211018

add a comment |

I would suggest you to write the PHP code outside otherwise script tag will be executed unnecessary with empty code if the condition is not met.

<?php 

    if(isset($usm , $ag) {

?>

<script>

    $( document ).ready(function() {

        var usmData = '<?php echo json_encode($usm); ?>';

        var agData = '<?php echo json_encode($ag); ?>';

    });

</script>

<?php

    }

?>

answered Jan 3 at 10:11

Rajendra arora

1,51211018

I would suggest you to write the PHP code outside otherwise script tag will be executed unnecessary with empty code if the condition is not met.

<?php 

    if(isset($usm , $ag) {

?>

<script>

    $( document ).ready(function() {

        var usmData = '<?php echo json_encode($usm); ?>';

        var agData = '<?php echo json_encode($ag); ?>';

    });

</script>

<?php

    }

?>

answered Jan 3 at 10:11

Rajendra arora

1,51211018

answered Jan 3 at 10:11

Rajendra arora

1,51211018

answered Jan 3 at 10:11

Rajendra arora

1,51211018

answered Jan 3 at 10:11

Rajendra arora

1,51211018

add a comment |

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