PHP preg_match - 如果匹配则回显结果？

Question

I'm trying to display hyperlinks to screen but only if a corresponding ID is recorded in a string.

<?php
function check($i, $s) {
    if (preg_match('/'.$i.'/',$test)) echo $s;
}

$test = "000,001,002,003,004,005";

check("001","<a href=''>This is in the test string - 001</a>");
check("003","<a href=''>This is in the test string - 003</a>");
check("006","<a href=''>This is in the test string - 006</a>");
check("020","<a href=''>This is in the test string - 020</a>");
?>

The desired output would be:

<a href=''>This is in the test string - 001</a>
<a href=''>This is in the test string - 003</a>

As they are the only two matches to the values in the string.

This isn't working.. Can you advise why and how to get it working.?

Thanks

Answer 1

You have not defined the variable $test in the function. You could do for example,

<?php
function check($i, $s, $test) {
    if (preg_match('/'.$i.'/',$test)) echo $s;
}

$test = "000,001,002,003,004,005";

check("001","<a href=''>This is in the test string - 001</a>", $test);
check("003","<a href=''>This is in the test string - 003</a>", $test);
check("006","<a href=''>This is in the test string - 006</a>", $test);
check("020","<a href=''>This is in the test string - 020</a>", $test);
?>

You should use error_reporting(E_ALL); when you develop, in that case you would have seen the following message:

Notice: Undefined variable: test in ... on line ...