Question

我一直得到相同的mysql错误代码，但我不知道如何纠正它。

错误：

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /websites/123reg/LinuxPackage22/we/ez/y_/weezy.co.uk/public_html/search.php on line 29

我在下面的代码中标出了第29行。

这是什么意思？

谢谢

<?php
// Change the fields below as per the requirements
$db_host="";
$db_username="";
$db_password="";
$db_name="";
$db_tb_name="data";
$db_tb_atr_name="name";
$db_tb_atr_name="email";
$db_tb_atr_name="location";


//Now we are going to write a script that will do search task
// leave the below fields as it is except while loop, which will display results on screen

mysql_connect("$db_host","$db_username","$db_password");
mysql_select_db("$db_name");

$query=mysql_real_escape_string($_GET['query']);

$query_for_result=mysql_query("

SELECT name, email, location

FROM data WHERE 

$db_tb_atr_name like '%".$query."%'");
echo "<h2>Search Results</h2><ol>";
while ($row = mysql_fetch_array($result))  <<<<<<<<<<<<<<< LINE 29 
{
echo "<li>";
echo substr($row["name"], $row["email"], $row["location"]);
echo "</li><hr/>";
}
echo "</ol>";

mysql_close();
?>

Answer 1

while ($row = mysql_fetch_array($result))

您将$result设置为何处？

你可能意味着

while ($row = mysql_fetch_array($query_for_result))

Answer 2

您将查询结果存储在名为$query_for_result的变量中，稍后您尝试阅读$result，这是空的。

Answer 3

您要将mysql_query的结果分配给$query_for_result，然后尝试在$result中循环mysql_fetch_array()。试试这个：

$result = mysql_query("...");
while ($row = mysql_fetch_array($result)) {
  ...

OR

$query_for_result = mysql_query("...");
while ($row = mysql_fetch_array($query_for_result)) {
  ...

Answer 4

您还应该检查是否有任何行返回

$rs = mysql_query("
  SELECT name, email, location
    FROM data 
   WHERE ".$db_tb_atr_name." like '%".$query."%'
");

if(mysql_num_rows($rs)){

  while ($row = mysql_fetch_array($rs)){
    // code
  }
} else {
// no results
}

调用mysql_fetch_array（）时获取“提供的参数不是有效的MySQL结果资源”

4 个答案: