troubleshooting Question
Code not seeing another code from other folders in project
Ivan Golubar asked on
I am working on a project and i am including code from different resources from the net. Now i have beginning to get problems. With "code not seeing another code ". I will explain on particular example:
My project tree is as next
In header.php i have a canvas <canvas class="objectcanvas" id="c" width="980" height="500" >
and in JSgetJson.js I am getting Json to display on canvas:
Until here everything works fine.
But am getting in troubles when i want to display the same Json in another.php on next canvas:
<canvas class="objectcanvas" id="c7" width="200" height="200" backgroundColor= "red" >
I have a problem alreday when i want to display canvas.
This is the error:
Uncaught ReferenceError: fabric is not defined at .....
Uncaught TypeError: Cannot create property 'style' on string 'c7' at i._createCanvasElement (fabric.min.js:2)
I am opening another.php page from the link in header.php.
And another.php is as next:
I hope that i gave enough information to start from somewhere.
My project tree is as next
project_folder
header.php
functions.php
PgetJson.php
js_folder
JSgetJson.js
fabric.min.js
sub_folder
another.php
In header.php i have a canvas <canvas class="objectcanvas" id="c" width="980" height="500" >
and in JSgetJson.js I am getting Json to display on canvas:
function getJsonF(){ //////////////////////////////////// JSgetjson
var whichProjectToSave=document.getElementById("selectProjectID").value;
$.ajax({
method:"POST",
url: '/wp-content/themes/net/PgJson.php',
data: {
"getCanvas":1,
"whichProject":whichProjectToSave,
},
datatype: "text",
success: function(strdate){
canvas.loadFromJSON(strdate, function() {
// console .log(strdate);
canvas.renderAll();
});
}
});
}
<?php ////////////////////////////////////////////////// PgetJson.php
$db2=mysqli_connect("localhost","ccc","ccc2","222cl_projectObjects")or die ("no connection");
if (isset($_POST["getCanvas"]) ) {
$projectName= mysqli_real_escape_string($db2, $_POST['whichProject']);
$query = "SELECT objectsList FROM projectObjectstable WHERE projectName='$projectName'";
$jsonCanvas= mysqli_query($db2,$query);
$row = mysqli_fetch_row($jsonCanvas);
$myLine=$row['0'];
echo $myLine;
}
mysqli_close($db2);
?>
Until here everything works fine.
But am getting in troubles when i want to display the same Json in another.php on next canvas:
<canvas class="objectcanvas" id="c7" width="200" height="200" backgroundColor= "red" >
I have a problem alreday when i want to display canvas.
This is the error:
Uncaught ReferenceError: fabric is not defined at .....
Uncaught TypeError: Cannot create property 'style' on string 'c7' at i._createCanvasElement (fabric.min.js:2)
I am opening another.php page from the link in header.php.
And another.php is as next:
<?php
...........
>
<!DOCTYPE html>
<html>
<head>
<style type="text/css">
....
.........
</style>
<script type="text/javascript">
.......
var canvas = new fabric.Canvas('c7');
...........
</script>
</head>
<body>
.......
........
<canvas class="objectcanvas" id="c7" width="100" height="100" backgroundColor= "red" >
........
............
</body>
</html>
I hope that i gave enough information to start from somewhere.
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