试图从文件中获取特定行。 程序跳过行。 （PHP）

I have a small problem. I need to get a line from a text file using PHP. Here is an example of the text file:

hello 2010-10-25
hello 2010-10-26
hello 2010-10-27
hello 2010-10-28
hello 2010-10-29
hello 2010-10-30
hello 2010-10-31

And my code for taking out a the line which contains "2010-10-26" is this:

<?php
 $datefile = fopen('file.txt', 'r') or exit("Unable to open file.txt");

 while(!feof($datefile))
 {
  $date = "2010-10-26";
  $string = fgets($datefile);
  if(strpos($string, $date)==true)
  {
   echo fgets($datefile);
  }

 }
 fclose($datefile);
?>

Instead of printing out the line "hello 2010-10-26", it prints out "hello 2010-10-27" I have no idea whats going on, please help.

When finding the row, you read the next line and return it.

echo fgets($datefile);

Instead you want to return the current line

echo $string;

Well like kingCrunch said instead of using

echo fgets($datefile);

use

echo $string;

or if you still want to use what you have been using, then set the date variable to a day less than what you want to get.

Hint: Its much more easier to use file() to read a lines of a file into an array.

Example:

$date = '2010-10-26';
foreach (file('file.txt') as $line) {
    if ($line == $date) {
        echo 'Match: '.$line;
        break;
    }
}

Or even shorter:

$date = '2010-10-26';
$lines = file('file.txt');

if (in_array('hello '.$date, $lines)) {
    echo 'Match '.$date;
}