Mixing
Index exception when using LIKE clause?
I'm calling mysqli_report(MYSQLI_REPORT_ALL) on my server to enable any warnings and errors that may need fixing. Since MYSQLI_REPORT_INDEX is inherited, PHP is also reporting bad/no index use.
I have set up my table properly, but still get the following exception when using a LIKE clause:
mysqli_sql_exception: No index used in query/prepared statement (null)
However, when using = instead of LIKE, I no longer get this error.
Is this fixable?
My table looks like this:
CREATE TABLE `articles` (
`ID` INT(10) UNSIGNED NOT NULL AUTO_INCREMENT,
`Title` VARCHAR(255) NOT NULL COLLATE 'utf8mb4_unicode_ci',
PRIMARY KEY (`ID`),
INDEX `Title` (`Title`)
);
And this is the code I'm testing on my server:
$db = mysqli_connect($Host, $User, $Pass, $Database);
$Stmt = mysqli_stmt_init($db);
// This works
mysqli_stmt_prepare($Stmt, 'SELECT ID FROM articles WHERE Title = ?');
$Title = 'test';
// This doesn't work
mysqli_stmt_prepare($Stmt, 'SELECT ID FROM articles WHERE Title LIKE ?');
$Title = '%test%';
mysqli_stmt_bind_param($Stmt, 's', $Title);
mysqli_stmt_execute($Stmt);
mysqli_stmt_bind_result($Stmt, $ID);
mysqli_stmt_fetch($Stmt);
mysqli_stmt_close($Stmt);
php mysql mysqli mariadb
edited Jan 1 at 22:47
GMB
17.7k3928
asked Jan 1 at 22:20
user966939user966939
496221
add a comment |
I'm calling mysqli_report(MYSQLI_REPORT_ALL) on my server to enable any warnings and errors that may need fixing. Since MYSQLI_REPORT_INDEX is inherited, PHP is also reporting bad/no index use.
I have set up my table properly, but still get the following exception when using a LIKE clause:
mysqli_sql_exception: No index used in query/prepared statement (null)
However, when using = instead of LIKE, I no longer get this error.
Is this fixable?
My table looks like this:
CREATE TABLE `articles` (
`ID` INT(10) UNSIGNED NOT NULL AUTO_INCREMENT,
`Title` VARCHAR(255) NOT NULL COLLATE 'utf8mb4_unicode_ci',
PRIMARY KEY (`ID`),
INDEX `Title` (`Title`)
);
And this is the code I'm testing on my server:
$db = mysqli_connect($Host, $User, $Pass, $Database);
$Stmt = mysqli_stmt_init($db);
// This works
mysqli_stmt_prepare($Stmt, 'SELECT ID FROM articles WHERE Title = ?');
$Title = 'test';
// This doesn't work
mysqli_stmt_prepare($Stmt, 'SELECT ID FROM articles WHERE Title LIKE ?');
$Title = '%test%';
mysqli_stmt_bind_param($Stmt, 's', $Title);
mysqli_stmt_execute($Stmt);
mysqli_stmt_bind_result($Stmt, $ID);
mysqli_stmt_fetch($Stmt);
mysqli_stmt_close($Stmt);
php mysql mysqli mariadb
edited Jan 1 at 22:47
GMB
17.7k3928
asked Jan 1 at 22:20
user966939user966939
496221
add a comment |
I'm calling mysqli_report(MYSQLI_REPORT_ALL) on my server to enable any warnings and errors that may need fixing. Since MYSQLI_REPORT_INDEX is inherited, PHP is also reporting bad/no index use.
I have set up my table properly, but still get the following exception when using a LIKE clause:
mysqli_sql_exception: No index used in query/prepared statement (null)
However, when using = instead of LIKE, I no longer get this error.
Is this fixable?
My table looks like this:
CREATE TABLE `articles` (
`ID` INT(10) UNSIGNED NOT NULL AUTO_INCREMENT,
`Title` VARCHAR(255) NOT NULL COLLATE 'utf8mb4_unicode_ci',
PRIMARY KEY (`ID`),
INDEX `Title` (`Title`)
);
And this is the code I'm testing on my server:
$db = mysqli_connect($Host, $User, $Pass, $Database);
$Stmt = mysqli_stmt_init($db);
// This works
mysqli_stmt_prepare($Stmt, 'SELECT ID FROM articles WHERE Title = ?');
$Title = 'test';
// This doesn't work
mysqli_stmt_prepare($Stmt, 'SELECT ID FROM articles WHERE Title LIKE ?');
$Title = '%test%';
mysqli_stmt_bind_param($Stmt, 's', $Title);
mysqli_stmt_execute($Stmt);
mysqli_stmt_bind_result($Stmt, $ID);
mysqli_stmt_fetch($Stmt);
mysqli_stmt_close($Stmt);
php mysql mysqli mariadb
edited Jan 1 at 22:47
GMB
17.7k3928
asked Jan 1 at 22:20
user966939user966939
496221
I'm calling mysqli_report(MYSQLI_REPORT_ALL) on my server to enable any warnings and errors that may need fixing. Since MYSQLI_REPORT_INDEX is inherited, PHP is also reporting bad/no index use.
I have set up my table properly, but still get the following exception when using a LIKE clause:
mysqli_sql_exception: No index used in query/prepared statement (null)
However, when using = instead of LIKE, I no longer get this error.
Is this fixable?
My table looks like this:
CREATE TABLE `articles` (
`ID` INT(10) UNSIGNED NOT NULL AUTO_INCREMENT,
`Title` VARCHAR(255) NOT NULL COLLATE 'utf8mb4_unicode_ci',
PRIMARY KEY (`ID`),
INDEX `Title` (`Title`)
);
And this is the code I'm testing on my server:
$db = mysqli_connect($Host, $User, $Pass, $Database);
$Stmt = mysqli_stmt_init($db);
// This works
mysqli_stmt_prepare($Stmt, 'SELECT ID FROM articles WHERE Title = ?');
$Title = 'test';
// This doesn't work
mysqli_stmt_prepare($Stmt, 'SELECT ID FROM articles WHERE Title LIKE ?');
$Title = '%test%';
mysqli_stmt_bind_param($Stmt, 's', $Title);
mysqli_stmt_execute($Stmt);
mysqli_stmt_bind_result($Stmt, $ID);
mysqli_stmt_fetch($Stmt);
mysqli_stmt_close($Stmt);
php mysql mysqli mariadb
php mysql mysqli mariadb
edited Jan 1 at 22:47
GMB
17.7k3928
asked Jan 1 at 22:20
user966939user966939
496221
edited Jan 1 at 22:47
GMB
17.7k3928
asked Jan 1 at 22:20
user966939user966939
496221
edited Jan 1 at 22:47
GMB
17.7k3928
edited Jan 1 at 22:47
GMB
17.7k3928
edited Jan 1 at 22:47
GMB
17.7k3928
17.7k3928
asked Jan 1 at 22:20
user966939user966939
496221
asked Jan 1 at 22:20
user966939user966939
496221
asked Jan 1 at 22:20
user966939user966939
496221
496221
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
This is a normal behavior.
mysql indexes work by indexing N characters starting from the left of the value.
As a consequence, your index is not used when you do :
WHERE Title LIKE '%Test%'
This is because the string to match starts with a wildcard, which represents a variable number of characters.
The index will be used if you do :
WHERE Title LIKE 'Test%'
Or of course :
WHERE Title = 'Test'
answered Jan 1 at 22:46
GMBGMB
17.7k3928
In most situations,Title LIKE 'Test'(no wildcard) is optimized toTitle = 'Test'.
– Rick James
Jan 2 at 2:23
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is a normal behavior.
mysql indexes work by indexing N characters starting from the left of the value.
As a consequence, your index is not used when you do :
WHERE Title LIKE '%Test%'
This is because the string to match starts with a wildcard, which represents a variable number of characters.
The index will be used if you do :
WHERE Title LIKE 'Test%'
Or of course :
WHERE Title = 'Test'
answered Jan 1 at 22:46
GMBGMB
17.7k3928
In most situations,Title LIKE 'Test'(no wildcard) is optimized toTitle = 'Test'.
– Rick James
Jan 2 at 2:23
add a comment |
This is a normal behavior.
mysql indexes work by indexing N characters starting from the left of the value.
As a consequence, your index is not used when you do :
WHERE Title LIKE '%Test%'
This is because the string to match starts with a wildcard, which represents a variable number of characters.
The index will be used if you do :
WHERE Title LIKE 'Test%'
Or of course :
WHERE Title = 'Test'
answered Jan 1 at 22:46
GMBGMB
17.7k3928
In most situations,Title LIKE 'Test'(no wildcard) is optimized toTitle = 'Test'.
– Rick James
Jan 2 at 2:23
add a comment |
This is a normal behavior.
mysql indexes work by indexing N characters starting from the left of the value.
As a consequence, your index is not used when you do :
WHERE Title LIKE '%Test%'
This is because the string to match starts with a wildcard, which represents a variable number of characters.
The index will be used if you do :
WHERE Title LIKE 'Test%'
Or of course :
WHERE Title = 'Test'
answered Jan 1 at 22:46
GMBGMB
17.7k3928
This is a normal behavior.
mysql indexes work by indexing N characters starting from the left of the value.
As a consequence, your index is not used when you do :
WHERE Title LIKE '%Test%'
This is because the string to match starts with a wildcard, which represents a variable number of characters.
The index will be used if you do :
WHERE Title LIKE 'Test%'
Or of course :
WHERE Title = 'Test'
answered Jan 1 at 22:46
GMBGMB
17.7k3928
answered Jan 1 at 22:46
GMBGMB
17.7k3928
answered Jan 1 at 22:46
GMBGMB
17.7k3928
answered Jan 1 at 22:46
GMBGMB
17.7k3928
17.7k3928
In most situations,Title LIKE 'Test'(no wildcard) is optimized toTitle = 'Test'.
– Rick James
Jan 2 at 2:23
add a comment |
In most situations,Title LIKE 'Test'(no wildcard) is optimized toTitle = 'Test'.
– Rick James
Jan 2 at 2:23
In most situations,
Title LIKE 'Test' (no wildcard) is optimized to Title = 'Test'.– Rick James
Jan 2 at 2:23
In most situations,
Title LIKE 'Test' (no wildcard) is optimized to Title = 'Test'.– Rick James
Jan 2 at 2:23
add a comment |
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