Question

我试图发布一个使用Jquery的值，所以我必须使用ajax将该值传递给php页面。但是当我这样做时，我得到了未定义的索引数据错误。

这是我的代码：

$("#requirementtable tbody tr").on("click", function(){
                        desc =  $(this).closest("tr").find("td:nth-child(4)").text().trim();
$.ajax({
        type:"POST",
        url:"get_diagnosis.php",
        data: desc,
        success: function (data, textStatus, jqXHR)
        {
            alert(data);
        },
        error: function (jqXHR, textStatus, errorThrown)
        {      
            alert("some error occured->" + jqXHR.responseJSON);
        }
    })
  })

这是我的PHP代码：

<?php 
    $data = $_POST['data'];
    echo $data;
?>

Answer 1

这是一个包含代码和解决方案的简单示例

//the data object     

         var obj={
                    name:"uneeb",
                    city:"gujranwala",
                    state: "punjab"
                };

    //the data object 



    //The ajax
        $.ajax({
                    type: 'post',
                    url: 'ajax/dynamic_update.php',
                    data:object,
                    success: function (data) {
                        location.reload();
                    }
                });
    //The ajax


the PHP part

echo $_POST['name']; //prints uneeb
echo $_POST['city']; //prints gujranwala
echo $_POST['state']; //prints punjab

Answer 2

检查desc变量中是否包含某个值，您必须post个数据，

data: {data:desc},

AJAX代码，

$.ajax({
    type:"POST",
    url:"get_diagnosis.php",
    data: {data:desc},
    success: function (data, textStatus, jqXHR) {
        alert(data);
    },
    error: function (jqXHR, textStatus, errorThrown) {      
        alert("some error occured->" + jqXHR.responseJSON);
    }
});

在PHP中使用isset()来防止未定义的错误，例如

<?php 
    $data = isset($_POST['data']) ? $_POST['data'] : 'Not defined';
    echo $data;
?>

如果您需要JSON响应格式，请在dataType:'json'中使用$.ajax，然后在PHP中使用json_encode()进行回复。

Answer 3

替换以下内容：

from Queue import PriorityQueue

def yield_chronologically(iterators):
    'iterators: list of iterator objects'
    PQ = PriorityQueue()

    # put first n items
    for i, it in enumerate(iterators):
        try:
            nxt = next(it)            
            # this is where you have to determine the priority
            # with a function get_chronological_key you have yet to write
            chronological_key = get_chronological_key(nxt) 
            PQ.put(chronological_key, (i, nxt))
        except StopIteration:
            pass

    # yield items and insert next item from iterator that was taken from
    # into the PQ
    while not PQ.empty():
        _, (i, nxt) = PQ.get()
        yield nxt
        try:
            nxt = next(iterators[i])                
            chronological_key = get_chronological_key(nxt)
            PQ.put(chronological_key, (i, nxt))
        except StopIteration:
            pass

并替换为：

desc =  $(this).closest("tr").find("td:nth-child(4)").text().trim();
//        remove ^^^^^^^^^^ since you're currently in tr click
desc =  $(this).find("td:nth-child(4)").text().trim();

使用：

data: desc,

在您的PHP代码中，您可以使用data: {data:desc}, //to pass data as object as you don't have serialized data获取数据。

Answer 4

将ajax中的data: desc,字段替换为data:{data: desc},，这是导致您无法在PHP代码中获取数据的原因

如何使用ajax从一个php页面发布值到另一个页面

4 个答案: