VBA(Access 2013)HTTP POST不传递参数

I am attempting to get my Access database written in VBA to communicate with a MySQL database on my clients website through a set of PHP web services I wrote. I have managed to get the Access db to retrieve data from the MySQL db but can't get it to post anything. I have narrowed the problem down to the fact that my HTTP request isn't sending the arguments I assign it.

Here are some questions and sites I have looked at already. These were not helpful, because the majority weren't dealing with PHP and were instead looking directly at websites, or were dealing with GET rather than POST.

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My code for VBA is:

Dim strJSONEncodedJob As String
        strJSONEncodedJob = "[{""ExpenseID"":""" & astrExpenseIDs(intI) & "}]"
        URL = "I removed the URL when posting"
        objHTTP.Open "POST", URL, False
        objHTTP.setRequestHeader "Content-Type", "application/json"
        objHTTP.send (strJSONEncodedJob)
        strResponse = objHTTP.responseText
        MsgBox strResponse

And my PHP code is:

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);

$json = file_get_contents('php://input');
$data = json_decode($json, true);

$stmt = $conn->prepare("DELETE FROM tblExpenses WHERE ExpenseID=?");
$txtExpenseID = $data['ExpenseID']; 
$stmt->bind_param("i", $txtExpenseID);

echo '{"result" : "success"}';


I get the success statement as a msgbox in VBA, as expected, however the record is not deleted from the MySQL db.

Does anyone have a solution?

Thanks in advance.

Update When I echo $json I get the JSON encoded string, which means the arguments ARE being passed. However, when I echo $data['ExpenseID'] I get a blank msgbox.

doumiyi7063 发表了答案......
5 年多之前 回复
duanchu3376 谢谢!索引做到了。这个问题一直在扼杀我,你修复了它。非常感谢。如何设置评论作为答案?
5 年多之前 回复
duanlu5055 你的json编码对象是一个包含单个对象的数组,所以如果你想获得该对象的属性,你首先必须从数组中索引它。类似于$txtExpenseID=$data[0]['ExpenseID'];(再次,我不是一个PHP人,所以只是猜测语法)
5 年多之前 回复
donglou8371 对不起这是一个错字,我会修改它。当我回显$json时,我得到了json字符串,这是我之前没有的,所以这表明参数正在被传递。但是我仍然无法执行该命令(删除MySQL中的记录)。当我回显$data['ExpenseID']时我什么都没得到。
5 年多之前 回复
donglanying3855 如果你回显$json和$txtExpenseID,你会看到你的期望吗?而且我不是PHP人员,但是在将它作为参数绑定之前,你不会填充$txtExpenseID的值吗?
5 年多之前 回复


你的json编码对象是一个包含单个对象的数组,所以如果你想获得该对象的属性,你首先要有 从数组中索引它。 类似</ p>

  $ txtExpenseID = $ data [0] ['ExpenseID'];

</ code> </ pre>

(同样,我不是一个PHP人员所以只是猜测语法)</ p>
</ div>



Your json-encoded object is an array containing a single object, so if you want to get at that object's properties you'd first have to index it from the array. Something like

$txtExpenseID = $data[0]['ExpenseID']; 

(again, I'm not a PHP person so just guessing the syntax)

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