CODEWITHSUNDEEP
javascriptphpjqueryarraysajax
Alex Parser
Alex Parser

Reputation: 43

Populating an array through jquery AJAX in php

I have a function in a compare.php that takes a parameter $data and uses that data to find certain things from web and extracts data and returns an array.

function populateTableA($data);

So to fill array I do this

$arrayTableA = populateTableA($name);

now this array is then used to iterate tables..

<table id="tableA">
<input type="text" name="search"/><input type="submit"/>
<?php foreach($arrayTableA as $row) { ?>
  <tr>
     <td><?php echo $row['name']?></td>
     <td><?php echo $row['place']?></td>
  </tr>
</table>

Now what I want to do is to enter some data on input and then through jquery ajax 

function populateTableA($data);

should be called and $array should be refilled with new contents and then populated on tableA without refreshing the page.

I wrote this jquery but no results.

$(document).on('submit',function(e) {
e.preventDefault();  // Add it here 
 $.ajax({ url: 'compare.php',
 var name = ('search').val();
     data: {action: 'populateTableA(name)'},
     type: 'post',
     success: function(output) {
                  $array = output;
              }
       });
});

I have been doing web scraping and the above was to understand how to implement that strategy... original function in my php file is below

function homeshoppingExtractor($homeshoppingSearch)
{
$homeshoppinghtml = file_get_contents('https://homeshopping.pk/search.php?category%5B%5D=&search_query='.$homeshoppingSearch);
$homeshoppingDoc = new DOMDocument();
libxml_use_internal_errors(TRUE); 
if(!empty($homeshoppinghtml)){
$homeshoppingDoc->loadHTML($homeshoppinghtml);
libxml_clear_errors(); 
$homeshoppingXPath = new DOMXPath($homeshoppingDoc);
//HomeShopping
$hsrow = $homeshoppingXPath->query('//a[@class=""]');
$hsrow2 = $homeshoppingXPath->query('//a[@class="price"]');
$hsrow3 = $homeshoppingXPath->query('(//a[@class="price"])//@href');
$hsrow4 = $homeshoppingXPath->query('(//img[@class="img-responsive imgcent"])//@src');

//HomeShopping
if($hsrow->length > 0){
    $rowarray = array();
    foreach($hsrow as $row){
        $rowarray[]= $row->nodeValue;
       // echo $row->nodeValue . "<br/>";
    }
}
if($hsrow2->length > 0){
    $row2array = array();
    foreach($hsrow2 as $row2){
        $row2array[]=$row2->nodeValue;
       // echo $row2->nodeValue . "<br/>";
    }
}
if($hsrow3->length > 0){
    $row3array = array();
    foreach($hsrow3 as $row3){
        $row3array[]=$row3->nodeValue;
        //echo $row3->nodeValue . "<br/>";
    }
}
if($hsrow4->length > 0){
    $row4array = array();
    foreach($hsrow4 as $row4){
        $row4array[]=$row4->nodeValue;
        //echo $row3->nodeValue . "<br/>";
    }
}
$hschecker = count($rowarray);
if($hschecker != 0) {
    $homeshopping = array();
    for($i=0; $i < count($rowarray); $i++){
        $homeshopping[$i] = [
        'name'=>$rowarray[$i],
        'price'=>$row2array[$i],
        'link'=>$row3array[$i],
        'image'=>$row4array[$i]
        ];
    }
}
else{
    echo "no result found at homeshopping";
}
}
return $homeshopping;
}

Upvotes: 1

Views: 86

Answers (3)

Aman Rawat
Aman Rawat

Reputation: 2625

You are in a very wrong direction my friend.

First of all there are some syntax error in your JS code.

So use JavaScript Debugging to find where you went wrong.

After that Basic PHP with AJAX to get a reference how ajax and PHP work together

Then at your code

  1. Create a PHP file where you have to print the table part which you want to refresh.
  2. Write an AJAX which will hit that PHP file and get the table structure from the server. So all the processing of data will be done by server AJAX is only used for request for the data and get the response from the server.
  3. Put the result in your html code using JS.

Hope this will help

Upvotes: 0

John C
John C

Reputation: 3112

As mentioned in the comments PHP is a server side language so you will be unable to run your PHP function from javascript.

However if you want to update tableA (without refreshing the whole page) you could create a new PHP page that will only create tableA and nothing else. Then you could use this ajax call (or something similar) -

$(document).on('submit','#formReviews',function(e) {
    e.preventDefault();
    $.ajax({ 
        url: 'getTableA.php', //or whatever you choose to call your new page
        data: {
            name: $('search').val()
        },
        type: 'post',
        success: function(output) {
            $('#tableA').replaceWith(output); //replace "tableA" with the id  of the table
        },
        error: function() {
            //report that an error occurred
        }
    });
});

Upvotes: 1

Sanooj T
Sanooj T

Reputation: 1327

Hi You are doing it in wrong way.You must change your response to html table and overwrite older one.

 success: function(output) {
              $("#tableA").html(output);
          }
   });

In your ajax page create a table with your result array

Upvotes: 0

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