Question

我使用jQuery和ajax調用了一個php頁面res.php。該代碼是： -

$('#submit_button').click(function() {   
    buildingVal = $("#building"); 
    levelVal = $("#level"); 
    data = 'building=' + buildingVal.val() + 'level=' + levelVal.val(); 
    $.ajax({ 

     url: "res.php", 
     type: "POST", 
     data: data,  
     success: function (data) { 
      $('#npc').html(data); 
     } 
    });​ 
}); 

的res.php頁面代碼： -

<?php 
//connect to the database 
$con = mysql_connect("localhost","root","12345") or die("error ".mysql_error()); 

//connect to the travian table 
mysql_select_db("trav",$con) or die("error ".mysql_error()); 

$building = mysql_real_escape_string($_GET['building']); 
$level = mysql_real_escape_string($_GET['level']); 


$query = "select * from "; 
$query = $query . $building; 
$query = $query . "where lvl=" . $level; 
$query = $query . ";"; 
$result = mysql_query($query) or die('Error in Child Table!'); 

$data = mysql_fetch_assoc($result); 

echo '<table><tr><td>Lumber=$data["lumber"]</td><td>Clay=$data["clay"]</td><td>Iron=$data["iron"]</td><td>Crop=$data["crop"]</td>'; 

?> 

我收到錯誤

Notice: Undefined index: building in C:\xampp\htdocs\debal\res.php on line 8 

Notice: Undefined index: level in C:\xampp\htdocs\debal\res.php on line 9 
Error in Child Table! 

我怎樣提取兩個是那樣的參數發送到頁面並在sql查詢中使用它們從數據庫中的表中檢索數據。 請你能幫助我..

Answer 1

問題是你通過AJAX發送$ _POST值，並嘗試從你的res.php頁面上的$ _GET分配變量。改變任何AJAX功能type: "GET"，或

$building = mysql_real_escape_string($_POST['building']); 
$level = mysql_real_escape_string($_POST['level']); 

上res.php

還你沒有正確發送 '等級' 變量

，你錯過了一個 '&'

data = 'building=' + buildingVal.val() + '&level=' + levelVal.val();