if (isset($_POST["getCanvas"])) {
$projectName= mysqli_real_escape_string($db2, $_POST['whichProject']);
$query = "SELECT objectsList FROM projectObjectstable WHERE projectName='$projectName'";
// $query = "SELECT objectsList,backgroundImage FROM projectObjectstable WHERE projectName='$projectName'";
$jsonCanvas= mysqli_query($db2,$query);
$row = mysqli_fetch_row($jsonCanvas);
$myLine=$row['0'];
echo $myLine;
}
使用上面的代码我从表中获得一列。我需要两列。我想尝试下一步:如何管理Ajax成功功能?
$query = "SELECT objectsList FROM projectObjectstable WHERE projectName='$projectName'";
$jsonCanvas= mysqli_query($db2,$query);
$row = mysqli_fetch_row($jsonCanvas);
$myLine=$row['0'];
echo $myLine;
$query2 = "SELECT backgroundImage FROM projectObjectstable WHERE projectName='$projectName'";
$jsonBackground= mysqli_query($db2,$query2);
$row2 = mysqli_fetch_row($jsonBackground);
$myLine2=$row['0'];
echo $myLine2;
}
对此,我需要下一个解决方案在这里。如何修改ajax成功函数在画布上获取两个变量(projectList和backgroundImage)?
$.ajax({
method:"POST",
url: '/wp-content/themes/mypage3/PgetJson.php',
data: {
"getCanvas":1,
"whichProject":whichProjectToSave
},
datatype: "text",
success: function(strdate){
canvas.loadFromJSON(strdate, function() {
canvas.renderAll();
});
}
});
一个额外的信息将不胜感激。在浏览器中调试php代码有哪些选择,因为它可以处理js? 谢谢
从了解json开始。 –
创建你的数据的数组,然后json_encode它并发送到ajax成功,并在那里解码并遍历它,并使用所有值 –
使用Ajax它是类型而不是方法 – Akintunde007