how to find a string within a string using php

Question

I am accessing a mysql database and displaying a column. in this column is a long string lets say its this:

<image identifier="540aa2ad-9a8d-454d-b915-605b884e76d5">
  <file><![CDATA[images/MV5BMTg5OTMxNzk4Nl5BMl5BanBnXkFtZTcwOTk1MjAwNQ@@._V1._SY317_CR0,0,214,317_.jpg]]></file>
  <title/>
  <link/>

whatever lies between <![CDATA[images/ and .jpg]]></file> will change on every row and i want to echo whatever lies between them pieces of code.

anyone help?

thanks

edit

so far i have:

function inStr ($needle, $haystack)
{
  $needlechars = strlen($needle); //gets the number of characters in our needle
  $i = 0;
  for($i=0; $i < strlen($haystack); $i++) //creates a loop for the number of characters in our haystack
  {
    if(substr($haystack, $i, $needlechars) == $needle) //checks to see if the needle is in this segment of the haystack
    {
      return TRUE; //if it is return true
    }
  }
  return FALSE; //if not, return false
}  
$img = '
SELECT *
FROM `item`
';
$result0 = mysql_query($img);

while ($row0 = mysql_fetch_array($result0)){

$haystack = $row0['elements'];
$needle = '<![CDATA[images/';
}

if(inStr($needle, $haystack))
{
  echo "string is present";
}

if the stuff that wraps around what you are searching for is consistent, substr with strpos is probably better. — ldg
– ldg, Commented Jun 11, 2011 at 17:13
I second regular expressions, though parsing it as an xml object is also an option. — datasage
– datasage, Commented Jun 11, 2011 at 17:15
@reece: minimally googling the function's name would yield the php docs -- complete with examples. — Denis de Bernardy
– Denis de Bernardy, Commented Jun 11, 2011 at 17:15
@denis i understand what the functions do, but i dont undestand how i can use them together to achieve what i want to achieve. which is why i asked on here to begin with. using strpos and substr so far i have got (see edit) — reece
– reece, Commented Jun 11, 2011 at 17:24

rid · Accepted Answer · 2011-06-11 17:36:48Z

1

$cdata_part = preg_quote('<![CDATA[images/');
$end_part = preg_quote('.jpg]]></file>');

if (preg_match("#{$cdata_part}(.+?){$end_part}#", $text, $matches)) {
    echo $matches[1];
}

answered Jun 11, 2011 at 17:36

rid

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Comments

Matthew · Accepted Answer · 2011-06-11 17:53:14Z

0

If you have XML, then it's easy:

<?php
$xml = <<<END
<image identifier="540aa2ad-9a8d-454d-b915-605b884e76d5">
  <file><![CDATA[images/MV5BMTg5OTMxNzk4Nl5BMl5BanBnXkFtZTcwOTk1MjAwNQ@@._V1._SY317_CR0,0,214,317_.jpg]]></file>
  <title/>
  <link/>
</image>
END;

$file = (string) simplexml_load_string($xml)->file;

echo $file;

You may need to adjust the simplexml "path" a bit if you only supplied us with a partial text blob. You make it sound like there are several images within the XML, but as I don't know it's formed I cannot really tell you how to iterate over it.

Oh, and this will return the .jpg portion too, but removing an extension is as simple as substr($file, 0, -4) if you know it is .jpg.

answered Jun 11, 2011 at 17:53

Matthew

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