我的php代码是:
$query = "SELECT * FROM users WHERE user='admin' AND password='MTIz'";
$result = $link->query($query);
$yes = array();
$yes[] = $result->num_rows;
echo json_encode($yes);
我的HTML代码是:
$.ajax({
url: 'vlogin.php',
type: 'POST',
data: myData,
dataType: 'json',
contentType: "application/json; charset=utf-8",
success: function(yes) {
alert(yes.Result);}
});
不返回任何东西。什么附录? 谢谢
使用 $.ajax() 时,我通常会实现错误回调:
来自 jquery 文档:error (类型: 函数( jqXHR jqXHR, 字符串文本状态, 字符串错误抛出 ))
实现它可以让您查看调用是生成(错误的 JSON)还是接收(网络服务错误)错误:您可以记录/警报文本状态和错误抛出。
$.ajax({
url: 'vlogin.php',
type: 'POST',
data: myData,
dataType: 'json',
contentType: "application/json; charset=utf-8",
success: function(yes) {
console.log(yes.Result);
},
error: function(jqXHR, textStatus, errorThrown) {
console.log(textStatus);
console.log(errorThrown);
}
});
$query = "SELECT * FROM users WHERE user='admin' AND password='MTIz'";
$result = $link->query($query);
$yes = array();
$yes[] = $result->num_rows;
echo json_encode($yes);
die;
在json_encode($yes);
后应用模具,它会给你结果,然后
并像这样只提醒是alert(yes);
试试下面
$query = "SELECT * FROM users WHERE user='admin' AND password='MTIz'";
$result = $link->query($query);
$yes = array();
$yes['record'] = $result->num_rows;
echo json_encode($yes);
而.ajax
$.ajax({
url: 'vlogin.php',
type: 'POST',
data: myData,
dataType: 'json',
contentType: "application/json; charset=utf-8",
success: function(yes) {
console.log(yes.record);
}
});