ajax-返回值 - 在使用ajax时,为什么使用get可以在后台php页面中获得值,而post确不行?
GET方式
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|
结果为:
string(4) "POST"
Notice: Undefined index: name1 in D:phpstudyWWWservice.php on line 23
Notice: Undefined index: number in D:phpstudyWWWservice.php on line 24
Notice: Undefined index: sex in D:phpstudyWWWservice.php on line 25
Notice: Undefined index: job in D:phpstudyWWWservice.php on line 26
{"success":"false","msg":"参数错误,填写信息不全"}
,这个是为什么呢?我哪里写的有问题?请高人指点,谢谢!
回复内容:
GET方式
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|
结果为:
string(4) "POST"
Notice: Undefined index: name1 in D:phpstudyWWWservice.php on line 23
Notice: Undefined index: number in D:phpstudyWWWservice.php on line 24
Notice: Undefined index: sex in D:phpstudyWWWservice.php on line 25
Notice: Undefined index: job in D:phpstudyWWWservice.php on line 26
{"success":"false","msg":"参数错误,填写信息不全"}
,这个是为什么呢?我哪里写的有问题?请高人指点,谢谢!
post 用$_POST取
如果你又想取get 还想取post 可以用 $_REQUEST
因为你 PHP 中是用$_GET 所以当然只能获取get传递过来的值
$_GET 获取 url连接中的参数
$_POST 获取 post提交的参数
$_REQUEST 可以获取到get和post的参数
你的案例里,用$_GET是获取不到Post参数的。
post请求时参数格式不正确,换成var postData = { "name1": "value1", "name2": "value2"} 这样试一试。可以F12调试一下。

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