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PHP string from an array element

Posted on 2007-03-21
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Last Modified: 2010-07-27
for ($i = 0; $i < count($lines); $i++)
{
if (strpos($lines[$i],"The river comes") >= 0)
{
      $current = $lines[$i];
}
}

This condition does show as true exactly once which is right.  $current, however,  is always just a return character.  I am getting so aggravated with this.
I have tried this using strval with the same result.
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Question by:okerupt
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3 Comments
 
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Accepted Solution

by:
Terry Woods earned 250 total points
ID: 18768789
When "The river comes" is at the start of the string, strpos returns 0 and $current is set. However, when it's not found at all, strpos returns false which seems to also evaluate to 0, so $current is also set (but it shouldn't be).

Try changing your test to:
  if (is_numeric(strpos($lines[$i],"The river comes")))
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LVL 35

Expert Comment

by:Terry Woods
ID: 18768793
Just so you know, I tested it as follows:

$lines = array("When The river comes by",
               "blah",
               "The river comes through");
for ($i = 0; $i < count($lines); $i++) {
  print "Searching $i: ".$lines[$i]."<br>\n";
  if (is_numeric(strpos($lines[$i],"The river comes"))) {
    $current = $lines[$i];
    print "Found: $current at pos ".strpos($lines[$i],"The river comes")."<br>\n";
  }
}

displays:

Searching 0: When The river comes by
Found: When The river comes by at pos 5
Searching 1: blah
Searching 2: The river comes through
Found: The river comes through at pos 0
0
 

Author Comment

by:okerupt
ID: 18770623
You're right about false = 0.  As a .net programmer that irritates me.
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