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Php SQL查询错误查询为空_Php_Mysql - Fatal编程技术网

Php SQL查询错误查询为空

Php SQL查询错误查询为空,php,mysql,Php,Mysql,您调用了两次mysql\u查询函数,但没有任何结果。尝试将您的线路替换为: <?php include "head-main.php"; include "conn.php"; if(isset($_GET)){ $var = $_GET["cat"]; } $data = mysql_query('SELECT kampword FROM data_dict where category=$var'); $retval = mysql_query

您调用了两次
mysql\u查询
函数,但没有任何结果。尝试将您的线路替换为:

<?php
    include "head-main.php";
    include "conn.php";

    if(isset($_GET)){ $var = $_GET["cat"]; }
    $data = mysql_query('SELECT kampword FROM data_dict where category=$var');

   $retval = mysql_query( $data, $conn );

     if(! $retval ) {
   die('Could not get data: ' . mysql_error());
   }

   while($row = mysql_fetch_assoc($retval)) {
   echo "World :{$row['kampword']}  <br> ".
     "--------------------------------<br>";
  }

  ?>

试一试


供参考。它们不再得到维护。看到了吗?相反,请了解,并使用or-将帮助您决定哪一个最适合您。您的脚本可能会有风险,您甚至可以查看使用时发生的情况。
$var
是字符串吗?它有价值吗?变量不会用简单的引号来解释。可能$data=mysql\u query(“从data\u dict中选择kampword,其中category=“.$var.”);错误已经消失,但仍然无法获取数据库中的数据
<?php
if(isset($_GET["cat"])) { 
    $var = $_GET["cat"];
    $retval = mysql_query('SELECT kampword FROM data_dict where category=$var');
}else {
    $retval = false;
}
$data = mysql_query("SELECT kampword FROM data_dict where category=$var");
$data = mysql_query('SELECT kampword FROM data_dict where category="'.$var.'"');
if(isset($_GET["cat"])){ $var = $_GET["cat"]; }