| Re: Requesting Help with a Regular Expression [message #179453 is a reply to message #179443] |
Mon, 29 October 2012 14:34  |
bobgatski
Messages: 11
Registered: October 2012
Karma:
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Junior Member |
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On Sunday, October 28, 2012 7:39:39 PM UTC-4, Denis McMahon wrote:
> On Sun, 28 Oct 2012 16:08:20 -0700, bobgatski wrote:
>
>
>
>> From strings such as these ...
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>> int(11)
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>> varchar(45)
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>> datetime ... I am trying to extract just the type and length; i.e. for
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>> the type I want ...
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>> int varchar datetime ... and for the length I want ...
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>> 11 45 NULL I can get the type using ...
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>> (\w+)
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>> ... but I haven't been able to get the type and length. What I think
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>> should work is ... (\w+)(\((\d+)\))?
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>> ... gets no matches at all,not even the type.
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>
>
> Try:
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>
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> preg_match( "/(\w+)\(?(\d+)?/", $target, $arr );
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>
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> <?php
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> $arr = array();
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> $arr[] = "int(11)";
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> $arr[] = "varchar(45)";
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> $arr[] = "datetime";
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>
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> foreach ( $arr as $x ) {
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> if ( preg_match( "/(\w+)\(?(\d+)?/", $x, $arr2 ) == 1 ) {
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> echo "\$x = \"{$x}\"";
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> echo ", \$arr2[0] = \"{$arr2[0]}\"";
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> echo ", \$arr2[1] = \"{$arr2[1]}\"";
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> if ( isset( $arr2[2] ) )
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> echo ", \$arr2[2] = \"{$arr2[2]}\"";
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> echo "\n";
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> }
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> else {
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> echo "\$x = \"{$x}\" - No match or regex error!\n";
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> }
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> }
Denis, Thank you very, VERY much. I'm very new to PHP so I don't fully understand all of your code, but I've tried it and it works and I am learning the language elements I need to fully understand it. Thanks again, Bob
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