嗨,下面的ajax将数据提交到URL。我已经在PHP文件中检索了数据,并且想编写一个SQL查询以选择WHERE数据等于已提交数据的数据。
见下文 -
function loadJobRequests() {
//AJAX code to submit form.
$.ajax({
type: "POST",
url: "http://localhost:8888/EduSubOct/json-data-
jobrequests.php",
data: { userEmail: localStorage.getItem("email")} ,
cache: false,
success: function(html) {
alert("Information Entered Successfully");
}
});
}
PHP SQL-
include_once'dbh.php';
//Here we fetch the data from the URL that was passed from our HTML
form
$userEmail = $_POST['userEmail'];
$sql = "SELECT * FROM jobRequest WHERE email='$userEmail'";
$result = mysqli_query($conn, $sql);
谁能告诉我这是否是对AJAX POST数据使用SELECT查询的正确方法。
最佳答案
从更改SQL语句
$sql = "SELECT * FROM jobRequest WHERE email='$userEmail'";
至
$sql = "SELECT * FROM jobRequest WHERE email='".$userEmail."'";
并且始终尝试检查该值是否已设置,有时没有发送任何值,并且您的php脚本将生成错误。所以你的整个代码是
if (isset( $_POST['userEmail'])) {
$userEmail = $_POST['userEmail'];
$sql = "SELECT * FROM jobRequest WHERE email='".$userEmail."'";
$result = mysqli_query($conn, $sql);
}