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HomeBackend DevelopmentPHP Tutorial请问ajax post表单,php怎么接收?

a.html
var params=$("#form1").serialize();
$.ajax({
    type: "post",
contentType: "application/text",
data:params,
dataType:"json",
//dataType: "binary",
        async: false,
timeout: 30000,
        url: '1.php',
        success: function(data){
                //处理data数据
var s = eval(data);
     alert(s.a+'  '+s.b);
        },
error: function(x, e) {
                   // alert(x.readyState);
alert(e);
                }
});
-------------------------------
1.php
$db = new mysqli('localhost','adb','winst','test');
mysqli_query($db, "set character set 'utf8'");
$arr = $_REQUEST;  
$vnameid =$arr['username'];
$vtele =$arr['telephone'];
$query = "INSERT INTO cardvip(namevip,tel) VALUES ('"+$vnameid+"','"+$vtele+"')";
$db->query($query);

$data = json_encode(array('a'=>$vnameid, 'b'=>$vtele)); 
echo $data;
?>

如果用$vnameid ='username';
$vtele ='telephone';
1.html alert(s.a+'  '+s.b);可正确显示。

就是接收不到post数据


回复讨论(解决方案)

注释掉 dataType:"json"
alert(s.a+'  '+s.b); 改成 alert(data);

1.php
print_r($_POST);
exit;

用此改动进行调试

注释掉 dataType:"json"
alert(s.a+'  '+s.b); 改成 alert(data);

1.php
print_r($_POST);
exit;

用此改动进行调试

返回数据没问题,
$arr = $_REQUEST;  
$vnameid =$arr['username'];
$vtele =$arr['telephone'];
我的意思是$vnameid 获得不到post数据

$vnameid =$arr['username'];取不到值?

$vnameid =$arr['username'];取不到值?
嗯,取不到值。
如果a.html中使用url: '1.php?'+params,
就可以取到值,这是怎么回事?

你的ajax参数dataType:"json",传递的却是serialize序列化的数据。
所以应该去掉dataType:"json"。

你按我说的做了吗?

你按我说的做了吗?

a.html
var params=$("#form1").serialize();
$.ajax({
    type: "post",
contentType: "application/text",
data:params,
        async: false,
timeout: 30000,
        url: '1.php',
        success: function(data){
                //处理data数据
alert(data);
        },
error: function(x, e) {
                   // alert(x.readyState);
alert(e);
                }
});
-----------------------------------
1.php
$arr = $_REQUEST;  
$vnameid =$arr['username'];
$vtele =$arr['telephone'];
echo $vnameid;
?>

还是获取不到post值

contentType: "application/text",
这个去掉吧。有这个MIME ?

删去 contentType: "application/text",

ok,去掉contentType: "application/text"就好了

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