Question

好吧，当我尝试使用MySQL信息填充下拉列表时，我遇到了问题。问题发生在第二次下拉时我试图从撇号的东西中获取信息......例如女士服装或男士服装。非常感谢任何帮助。

以下是错误：您的SQL语法出错;检查与MySQL服务器版本对应的手册，以便在第1行的“服装”附近使用正确的语法

这是代码。

<?php
//**************************************
//     Page load dropdown results     //
//**************************************
function getTierOne()
{
    $catresult = mysql_query("SELECT DISTINCT category FROM categories") 
    or die(mysql_error());

      while($tier = mysql_fetch_array( $catresult )) 

        {
           echo '<option value="'.$tier['category'].'">'.$tier['category'].'</option>';
        }

}

//**************************************
//     First selection results     //
//**************************************
if($_GET['func'] == "drop_1" && isset($_GET['func'])) { 
   drop_1($_GET['drop_var']); 
}

function drop_1($drop_var)
{  
    include_once('db.php');
    $result = mysql_query("SELECT DISTINCT level1 FROM categories WHERE category='$drop_var'") 
    or die(mysql_error());

    echo '<select name="drop_2" id="drop_2">
          <option value=" " disabled="disabled" selected="selected">Choose one</option>';

           while($drop_2 = mysql_fetch_array( $result )) 
            {
              echo '<option value="'.$drop_2['level1'].'">'.$drop_2['level1'].'</option>';
            }

    echo '</select>';
    echo "<script type=\"text/javascript\">
$('#wait_2').hide();
    $('#drop_2').change(function(){
      $('#wait_2').show();
      $('#result_2').hide();
      $.get(\"pla2.php\", {
        func: \"drop_2\",
        drop_var: $('#drop_2').val()
      }, function(response){
        $('#result_2').fadeOut();
        setTimeout(\"finishAjax_tier_three('result_2', '\"+escape(response)+\"')\", 400);
      });
        return false;
    });
</script>";
}


//**************************************
//     Second selection results     //
//**************************************
if($_GET['func'] == "drop_2" && isset($_GET['func'])) { 
   drop_2($_GET['drop_var']); 
}

function drop_2($drop_var)
{  
    include_once('db.php');
    $bresult = mysql_query("SELECT DISTINCT level2 FROM categories WHERE level1='$drop_var'") 
    or die(mysql_error());

    echo '<select name="drop_3" id="drop_3">
          <option value=" " disabled="disabled" selected="selected">Choose one</option>';

           while($drop_3 = mysql_fetch_array( $bresult )) 
            {
              echo '<option value="'.$drop_3['level2'].'">'.$drop_3['level2'].'</option>';
            }

    echo '</select> ';
    echo '<input type="submit" name="submit" value="Submit" />';
}
?>

Answer 1

使用htmlentities？

.htmlenteties($drop_2['level2']).

Answer 2

您可以将其替换为html特殊字符，替换为单引号字符。请参阅list here：

Answer 3

我明白了。我不得不改变

"SELECT DISTINCT level2 FROM categories WHERE level1='$drop_var'"

到

"SELECT DISTINCT level2 FROM categories WHERE level1='".mysql_real_escape_string$drop_var"'"

动态下拉mysql撇号问题

3 个答案: