JQuery Ajax沒有更新記錄

Question

我有一些jquery ajax代碼來更新記錄，但它不像預期的那樣工作。

下面是代碼：

function update_records() { 
    $.ajax({ 
    type: "POST", // Set the type then $_GET['id'] or $_POST['id'] 
    url: "update_record.php", 
    data: { category: "John", image: "Boston" } 
}).done(function(msg) { 
    alert("Data Saved: " + msg); 
}); 

}; //end function 


Then the php 

<?php 

    $id = $_REQUEST['id']; 


    include 'config.php'; 


    $result = mysql_query("UPDATE mytable SET title = 'something' where id=$id"); 
    $result = mysql_query("SELECT * FROM mytable WHERE id = '$id'");    
    $array = mysql_fetch_row($result); 

?> 

Answer 1

的例子我能不明白你的查詢返回。 也許你需要選擇其中id = $id;

// $id = $_POST['id'] or $_GET['id'] Where is $id??? 
    $result = mysql_query("UPDATE mytable SET title = 'something' where id=$id"); 
    $result = mysql_query("SELECT * FROM mytable WHERE id = '$id'");    
    $array = mysql_fetch_row($result);  


//You can use mysql_error() function to see the error. 
$result = mysql_query("UPDATE mytable SET title = 'something' where id=$id") or die(mysql_error()); 

Answer 2

u需要

$id = $_REQUEST['id']; 

我想你忘記執行更新查詢前加入這一行。從jQuery.com http://api.jquery.com/jQuery.ajax/

$.ajax({ 
    type: "POST", // Set the type then $_GET['id'] or $_POST['id'] 
    url: "some.php", 
    data: { name: "John", location: "Boston" } 
}).done(function(msg) { 
    alert("Data Saved: " + msg); 
}); 

Answer 3

我看你有：

$result = mysql_query("UPDATE mytable SET title = 'something' where id=$id"); 

我沒有看到你拿的$id