Heim >Backend-Entwicklung >PHP-Tutorial >ajax post数据提交 PHP后台接收不到数据
JS的脚本提交
$.ajax({ url:"/music/json.php", type:"POST", data:{"id":id}, dataType:"json", success: function(e){
//这里不知道怎么接收数据 echo $_POST["id"]; //不行 echo json_decode($_POST['data']);//不行//下面就是数据库查询$query="select * from gequ where id='$_POST[id]'";//执行查询语句不正确,尝试各种方法如上面获取ID的数据未果来求助$result=mysql_query( $query );
正常提交请不要加dataType:"json",
否则你需要print_r(json_decode($_POST['id']))这个是数组 不能echo
刚写了ajax post删除 给你发出来吧
function deleteall(){ var checkboxs = document.getElementsByName("selects"); var ids = new Array(); for(var i = 0 ; i<checkboxs.length ; i++){ if(checkboxs[i].checked == true){ ids[i] = checkboxs[i].value; } } if(ids.length<1){ return false; } var newids = ids.join(","); if(confirm("确认删除?")){ $.ajax({ type: "POST", url: "<?php echo url('autoReply','deleteall')?>", data:"id="+newids, success: function(e){ if(e == 1 ){ for(var i = 0 ; i<checkboxs.length ; i++){ if(checkboxs[i].checked == true){ checkboxs[i].checked = false; } } location.reload(); }else{ alert("删除失败"); } } }) } }
另外
$query="select * from gequ where id='$_POST[id]'";
data:{"id":id},
在JS里面alert一下这个id,看是否有值
<html> <head> <title>title</title> <script src="http://libs.baidu.com/jquery/1.9.0/jquery.min.js"></script> <script> $(function(){ $("#btn").click(function(){ alert('ok!'); var id = 3; $.ajax({ url:"2.php", method:"POST", dataType:"json", data:{id:id}, success:function(data){ alert(data.id); } }); }); }); </script> </head> <body> <input type="button" id="btn" value="button"> </body></html>
<?php$dd['id'] = $_POST['id'];$dd['value'] = 'fff';echo json_encode($dd);?>
dataType (default: Intelligent Guess (xml, json, script, or html))
Type: String
The type of data that you're expecting back from the server. I
正常提交请不要加dataType:"json",
否则你需要print_r(json_decode($_POST['id']))这个是数组 不能echo
我擦 楼主你看看5楼,
你看看 官方给的 介绍:http://api.jquery.com/jquery.ajax/
dataType (default: Intelligent Guess (xml, json, script, or html))
Type: String
The type of data that you're expecting back from the server. If none is specified, jQuery will try to infer it based on the MIME type of the response (an XML MIME type will yield XML, in 1.4 JSON will yield a JavaScript object, in 1.4 script will execute the script, and anything else will be returned as a string). The available types (and the result passed as the first argument to your success callback) are:
<html> <head> <title>title</title> <script src="http://libs.baidu.com/jquery/1.9.0/jquery.min.js"></script> <script> $(function(){ $("#btn").click(function(){ alert('ok!'); var id = 3; $.ajax({ url:"2.php", method:"POST", dataType:"json", data:{id:id}, success:function(data){ alert(data.id); } }); }); }); </script> </head> <body> <input type="button" id="btn" value="button"> </body></html>
<?php$dd['id'] = $_POST['id'];$dd['value'] = 'fff';echo json_encode($dd);?>
dataType (default: Intelligent Guess (xml, json, script, or html))
Type: String
The type of data that you're expecting back from the server. I
我擦 楼主你看看5楼,
你看看 官方给的 介绍:http://api.jquery.com/jquery.ajax/
dataType (default: Intelligent Guess (xml, json, script, or html))
Type: String
The type of data that you're expecting back from the server. If none is specified, jQuery will try to infer it based on the MIME type of the response (an XML MIME type will yield XML, in 1.4 JSON will yield a JavaScript object, in 1.4 script will execute the script, and anything else will be returned as a string). The available types (and the result passed as the first argument to your success callback) are:
我说的datatype不用写是如果不牵扯到跨域传值或者js的数组 /对象 的话 字符串完全可解决 所以就没必要加datatype 比如你要已json格式传递
另外 url:"/music/", 很简单就能理解 就是当前目录下的/music/
比如你访问www.xxx.com/music/ 那么如果设置了默认文件了 会直接访问默认文件的
比如默认文件时index.php那么实际就是www.xxx.com/music/index.php
另外json_decode()有2个参数 第一个是json 第二个如果不想要对象的话请写true