[英]how can i send a variable to php file
can someone show me how to send an variable to my php file out of that construct?有人可以告诉我如何从该构造中将变量发送到我的 php 文件吗? i am new at ajax/javascript and i try to figure out my style for writing JS code.
我是 ajax/javascript 的新手,我试图找出我编写 JS 代码的风格。 atm i like and understand that kind of version.
atm 我喜欢并理解那种版本。
var ajax = new XMLHttpRequest();
var method = "POST"
var url = "data.php" //erweitern - später
var asynchronous = true
var button = document.getElementById("count")
var counter = 0;
button.onclick = function() {
counter += 2;
showMoreFood(counter)
};
ajax.open(method, url, asynchronous)
ajax.send()
ajax.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
var data = JSON.parse(this.responseText)
var html =""
//for (var a = 0; a < data.length; a++) {
for (var a = 0; a < 2; a++) {
var firstname = data[a].name
var preis = data[a].preis
html += "<tr>"
html += "<td>" + firstname + "</td>"
html += "<td>" + preis + "</td>"
html += "</tr>"
}
document.getElementById("data").innerHTML = html
} // if
} // onready
what i tried:我试过的:
var data = "variable1"
ajax.open(method, url, asynchronous, data)
in data.php
-----------
$variableA = $_POST['variable1'];
because i thought i could send the data within, but nope.因为我以为我可以在其中发送数据,但是不行。 thx and br
thx 和 br
i got it.我知道了。 i just need to change that to
我只需要把它改成
method = "GET"
url = "data.php?variable1=something"
and in the php file并在php文件中
$variableA = $_GET['variable1'];
and it works :) i dont think thats the right version to handle that but it works :)它有效:) 我不认为那是处理它的正确版本,但它有效:)
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