我试图在PHP中创建一个更新函数,但是记录似乎并没有随着更新而改变。我创建了一个JSON对象来存放传递给这个文件的值,并且根据我运行的Firebug Lite控制台,这些值被输出得很好,所以它在sql方面可能有些问题。任何人都可以发现问题吗?我会很感激帮助!在PHP中不更新的SQL表格
<?php
$var1 = $_REQUEST['action']; // We dont need action for this tutorial, but in a complex code you need a way to determine ajax action nature
$jsonObject = json_decode($_REQUEST['outputJSON']); // Decode JSON object into readable PHP object
$name = $jsonObject->{'name'}; // Get name from object
$desc = $jsonObject->{'desc'}; // Get desc from object
$did = $jsonObject->{'did'};// Get id object
mysql_connect("localhost","root",""); // Conect to mysql, first parameter is location, second is mysql username and a third one is a mysql password
@mysql_select_db("findadeal") or die("Unable to select database"); // Connect to database called test
$query = "UPDATE deal SET dname = {'$name'}, desc={'$desc'} WHERE dealid = {'$did'}";
$add = mysql_query($query);
$num = mysql_num_rows($add);
if($num != 0) {
echo "true";
} else {
echo "false";
}
?>
请注意'{'$ name'}'和''{$ name}''之间有区别。也许你的WHERE条件不正确。 – EmCo 2013-02-19 18:11:29