在php中使用JQ的get json
js code
<script>
$(function() {
var chkDataURL= "API//chkIsSameData.php?name=alonso";
var jqxhr = $.getJSON( chkDataURL, function(jsonresult) {
debugger;
console.log(jsonresult);
})
});
</script>
/////php code
<?php require_once('連線字串的檔案.php'); ?>
<?php
if(isset($_GET['name']) ){
$name = $_GET['name'];
$sql = "select * from members where name = '".$name ."'";
$result = mysqli_query($conn, $sql ) or die(mysqli_error($conn));
$rows = mysqli_num_rows($result);
$response = array();
if($rows > 0){
$response =array("Y");
}
echo json_encode($response); //輸出的格式必須是json
}
?>
以上文章僅用紀錄資料使用.....