初学ajax和php不是特别会
我的逻辑是点击compid的这个点,讲道理应该会弹出alert,但点完之后就是没有响应
ajax的部分
$.ajax({
url : "first.php",
type : "post",
data : {
"companyId" : compId
},
dataType : "json",
success : function(result) {
alert(result.x);
}
});
php部分
<?php
header("Content-Type: text/html;charset=utf-8");
$companyId = isset($_POST["companyId"]) ? $_POST["companyId"] : "";
$con = mysqli_connect('localhost','root','');
if (!$con)
{
die('Could not connect: ' . mysqli_error($con));
}
// 选择数据库
mysqli_select_db($con,"gis_ks");
// 设置编码,防止中文乱码
mysqli_set_charset($con, "utf8");
$sql="SELECT * FROM company WHERE id = '".$companyId."'";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result)){
$arr[]= array(
'id' => $row['id'],
'name' => $row['name'],
'x' => $row['x'],
'y' => $row['y'],
'province' => $row['province'],
);
}
echo json_encode($arr);
mysqli_close($con);
?>