您好我尝试从这个
的js文件中的php链接加载JSON数据$。getJSON('link to rempl.php',function(data){
cn = data.nom;
});
这是rempl.php的代码
<?php
header('Content-Type: text/plain; charset=utf-8');
require_once ('connectDB.php');
if ($_POST['nom']== 0){
header("Location: http://localhost/geopp/editer.php");
}else{
$id = $_POST['nom'];
}
$sql = "SELECT id , id_archi , nom , archithect , adresse , date_construction , dateconst , resume FROM patri where id=".$id.";";
$req = mysqli_query ($link,$sql);
$feature = array();
//echo "lon\tlat\ttitle\tdescription\ticon";
while ($row = mysqli_fetch_assoc($req)) {
$res['id_archi'] = $row['id_archi'];
$res['id'] = $row['id'];
$res['date'] = $row['dateconst'];
$res['nom'] = $row['nom'];
$res['archithect'] = $row['archithect'];
$res['adresse'] = $row['adresse'];
$res['date_construction'] = $row['date_construction'];
$res['resume'] = $row['resume'];
$feature[] = json_encode($res);
}
echo implode(', ',$feature);
header("Location: http://localhost/geopp/empl.php");
?>
但当我警告变量cn时,它显示我未定义并感谢帮助我。