Get url parameter and query mysql data with ajax - php

Question

I want to get a parameter from an url. The url looks like this:
www.example.com/?v=12345
I want to get the parameter and query my mysql database to get the right data with ajax.
So i have my ajax call here:
$.ajax({
type:"POST",
url:"ajax2.php",
dataType:"json",
success:function(response){
var id = response['id'];
var url = response['url'];
var name = response['name'];
var image = response['image'];
},
error:function(response){
alert("error occurred");
}
});
As you can see, the data which i want to get are in a json array and will be saved in javascript variables.
This is my php file:
<?php
// Connection stuff right here
$myquery = "SELECT * FROM mytable **WHERE id= **$myurlvariable**;
$result = mysql_query($myquery);
while($row = mysql_fetch_object($result))
{
$currentid = "$row->id";
$currentname = "$row->name";
$currenturl = "$row->url";
$currentimage = "$row->image";
$array = array('id'=>$currentid,'url'=>$currenturl, 'name'=>$currentname,'image'=>$currentimage);
echo json_encode($array);
}
?>
The part where i want to query the right variable is bolded. I don't know how to query that. And Furthermore how to even get the url parameter in the proper form.
Can anybody help? Thank you!

Answer 1

You can get the query string using JavaScript and send it in the AJAX request.
Getting the query string(JavaScript) -
function query_string(variable)
{
var query = window.location.search.substring(1);
var vars = query.split("&");
for (var i=0;i<vars.length;i++) {
var pair = vars[i].split("=");
if(pair[0] == variable){return pair[1];}
}
return(false);
}
//Getting the parameter-
v = query_string('v'); // Will return '12345' if url is www.example.com/?v=12345
This needs to be passed as data in the AJAX call.
$.ajax(
{
type: "POST",
dataType: "json",
url: "ajax2.php",
data: "v="+v,
success: function(response){
var id = response['id'];
var url = response['url'];
var name = response['name'];
var image = response['image'];
},
error: function(jqXHR,textStatus,errorThrown){
//alert(JSON.stringify(jqXHR));
//alert(textStatus);
//alert(errorThrown);
alert(JSON.stringify(jqXHR)+" "+textStatus+" "+errorThrown);
//alert("error occurred");
}
}
);
This can be accessed as $_POST['v'] in the php form.
if(isset($_POST['v'])){
$myurlvariable = $_POST['v'];
$myquery = "SELECT * FROM mytable WHERE id= $myurlvariable";
...
And in php form, before you echo out the json response, change the content type. Something like this-
header("Content-Type: application/json");
echo json_encode($array);
If there is a database error, then it has to be handled.
So do this -
<?php
// Connection stuff right here
header("Content-Type: application/json");
if(isset($_POST['v'])){
$myurlvariable = $_POST['v'];
$myquery = "SELECT * FROM mytable WHERE id= $myurlvariable";
$result = mysql_query($myquery) or die(json_encode(Array("error": mysql_error()));
while($row = mysql_fetch_object($result))
{
$currentid = "$row->id";
$currentname = "$row->name";
$currenturl = "$row->url";
$currentimage = "$row->image";
$array[]= array('id'=>$currentid,'url'=>$currenturl, 'name'=>$currentname,'image'=>$currentimage);
}
echo json_encode($array);
}else{
echo json_encode(Array("error": "No POST values"));
}
?>
So this way, if the query has not executed properly, then you will know what exactly the error is.

Answer 2

Without any error checking, just the important part:
$myquery = "SELECT * FROM mytable WHERE id=" . $_POST['v'];