Question

我已经使用以下新错误编写了此帖子

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in E:\webareas\ie803\projectx\search.php on line 306

我的php代码如下：

    <?php

$car = mysql_real_escape_string($_REQUEST['car']);
$model = mysql_real_escape_string($_REQUEST['model']);
$type = mysql_real_escape_string($_REQUEST['type']);
$colour = mysql_real_escape_string($_REQUEST['colour']);
$year = mysql_real_escape_string($_REQUEST['year']);
$price = mysql_real_escape_string($_REQUEST['price']);


$con = mysql_connect("--","---","---");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db('-----', $con);

$sql = "SELECT * FROM Cars WHERE Make ='$car', Model ='$model', Type ='$type', Colour = '$colour', Year  = '$year', Price = '$price'";
$result=mysql_query($sql, $con);


{
     while($info = mysql_fetch_array($result)){
        echo "<tr>";
        echo "<td>" . $info['Make']. "</td>";
        echo "<td>" . $info['Model']. "</td>";
        echo "<td>" . $info['Type']. "</td>";
        echo "<td>" . $info['Colour']. "</td>";
        echo "<td>" . $info['Year']. "</td>";
        echo "<td>" . $info['Price']. "</td>";


        echo "<br/><br/><td>" . '<hr>' .  "</td>";


}
}
echo "</tr>";
echo "</table>";
?>

第306行是while语句。我每次改变后都会继续犯错误。

非常感谢

Answer 1

select语句只是一个字符串，你必须通过mysql_query提供它来获取结果集，它通常按如下方式完成：

$sql = "SELECT * FROM Cars WHERE Make ='$car', Model ='$model', Type ='$type', Colour = '$colour', Year  = '$year', Price = '$price'";
$result=mysql_query($sql, $con);

Answer 2

您没有mysql_query（），请尝试

$sql = "SELECT * FROM Cars WHERE Make ='$car', Model ='$model', Type ='$type', Colour = '$colour', Year  = '$year', Price = '$price'";
$result = mysql_query($sql) or die("ERROR: ".mysql_error());

我希望有帮助

修改我输入或者使用die语句来帮助调试，你可以随时删除它。我个人觉得它更容易。

Answer 3

您需要在mysql_query之前使用mysql-fetch-aarray。请参阅此page，了解如何执行此操作

Answer 4

我认为你的查询甚至没有被执行。我以前从未见过WHERE [column], [column], [column]的语法。

尝试：

SELECT * FROM Cars WHERE Make ='$car' AND Model ='$model' AND Type ='$type' AND Colour = '$colour' AND Year  = '$year' AND Price = '$price'

我添加了一些错误检查，以便您可以看到所获得的任何mysql错误，如果语法正确的话。

$sql = "SELECT * FROM Cars WHERE Make ='$car', Model ='$model', Type ='$type', Colour = '$colour', Year  = '$year', Price = '$price'";
$result=mysql_query($sql, $con);
if($result)    
{
     while($info = mysql_fetch_array($result)){
        echo "<tr>";
        echo "<td>" . $info['Make']. "</td>";
        echo "<td>" . $info['Model']. "</td>";
        echo "<td>" . $info['Type']. "</td>";
        echo "<td>" . $info['Colour']. "</td>";
        echo "<td>" . $info['Year']. "</td>";
        echo "<td>" . $info['Price']. "</td>";


        echo "<br/><br/><td>" . '<hr>' .  "</td>";


}
else {
    die(mysql_error();
}

选择声明不起作用

4 个答案: