我需要将数据js发送到php 但是将变量javascript传递给php时出现问题。 我一直在寻找Stackoverflow中的任何问题,但是我认为没有像我这样的问题。
function test() {
var xhttp;
var text1 = "nasjdnknsak";
xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function () {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("txtHint").innerHTML = this.responseText;
}
};
xhttp.open("POST", "train.php", true);
xhttp.send("text1=dfsbbdfsknkfj");
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "saravine";
$text1 = $_POST['text1'];
die("Masuk ga : " . $text1);
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO train (caption)
VALUES ('$text1')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();