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;
}
使用上面的代碼我從表中獲得一列。我需要兩列。我想嘗試下一步: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;
}
對此,我需要下一個解決方案在這裏。如何修改ajax成功函數在畫布上獲取兩個變量(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();
});
}
});
一個額外的信息將不勝感激。在瀏覽器中調試php代碼有哪些選擇,因爲它可以處理js? 謝謝
從瞭解json開始。 –
創建你的數據的數組,然後json_encode它併發送到ajax成功,並在那裏解碼並遍歷它,並使用所有值 –
使用Ajax它是類型而不是方法 – Akintunde007