Question

i am trying to do a guestbook in php but i am having some problems with mysql_fetch_array function. I don t understand why. I try to debug by putting die("Error ".mysql_error()) but nothing prints out. I guarantee that all my variables are correctly initialized. Here is my code :

<?php

 $nbmessagesPP = 10;
 mysql_connect(HOST, USER,PASSWORD) or die( "Unable to connect to database");
 mysql_select_db(DBNAME) or die ("Unable to select database!"); 

 .......

 if(isset($_GET[ page ])){
    $page = $_GET[ page ];
  } else {
    $page = 1;
  }
  $first_msg = ($page - 1) * $nb_of_Page;
  $query =  Select * from livredor ORDER BY id DESC LIMIT  .$first_msg. ,  .$nbmessagesPP;
  $rep = mysql_query($query) or exit("Error in query".mysql_error());

  $v = true;
  while($v){
      $v = ($data = mysql_fetch_array($rep) or die ("Error fetching the data : ".mysql_error()));
      echo "<p>id -> ".$data[ id ]."</p>";    
      echo "<p>pseudo ->".$data[ pseudo ]."</p>";
      echo "<p>messages ->".$data[ message ]."</p>";
      echo "<hr/>";
  } 
  mysql_close();
?>

一个人能够帮助我;

Answer 1

你的法典没有正确地处理错误或最后一行。当v 是假的时,它仍在印刷一些数据。最好改写如下:

while (($data = mysql_fetch_array($rep))) {    
  echo   
  ... 
}

这需要在开始印刷之前对单体进行评价。

Answer 2

问题在于,你再次试图获取没有结果的部分。 mysql_fetch_array> 回归一个固定阵列,并附有立体指数。您想要的是mysql_fetch_assoc,该编码回归一个联系阵列。

<><>Edit>: 你们也有克里斯特的问题,没有正确处理最后一行。

Answer 3

一般来说,如果你再次收到一个错误的话,即“应用论点不是一种有效的MySQL结果资源”,就意味着MySQL的查询失败了,那就没有归还有效的结果资源。

在通过<代码>mysql_query(<>/code>)发送之前,请填上$query,然后尝试将重复的质询带入phpMyAdmin,看看它是否恢复任何结果。

Answer 4

Ok i have found the problem. The problem was that in another page i had a mysql_connection and in that page i was creating a new one. I just catch the return value of mysql_connect function and then close it with mysql_close function at the end. Like this :

  <?php
     $link = mysql_connect(HOST, USER,PASSWORD) or die( "Unable to connect to database");
     mysql_select_db(DBNAME) or die ("Unable to select database!"); 
     .....

     while($data = mysql_fetch_array($rep)) {//i do something here}

     mysql_close($link);    
 ?>

感谢您的答复:

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