Passing data from one PHP page to another PHP Page

Question

I have a program that I'm making that takes a database gets the top 5 rows from that database and presents it to a user when a button is clicked on the website I have made. I was able to get the database data to print on the page where the button was clicked, but I want it to be printed to another page that I have created. Here is my code so far for the main page with the buttons and the PHP code to get the database information(the file is called file1.php):

<html>
    <head>

        <title>Testing SQL DB</title>
        <link rel="stylesheet" href="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css">
        <link rel="stylesheet" href="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.5/css/bootstrap-theme.min.css">
        <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>


    <script type = "text/javascript">
        function myAjax () {
            $.ajax( { type : 'POST',
              data : { },
              url  : 'query.php',              // <=== CALL THE PHP FUNCTION HERE.
              success: function ( data ) {
                document.getElementById("dbResult").innerHTML = data;   // Get the element with id="demo"
              },
              error: function ( xhr ) {
                alert( "error" );
              }
            });
        }
    </script>
    </head>
    <style>
        h1
        {
            text-align: center;
            font-size: 45px;
        }

        .radio-input
        {
            font-size: 35px;
        }
        .radio
        {
            font-size: 45px;
            display: block;
            vertical-align: top;
            margin-right: 3%;
            margin-bottom: 30%;
            text-align: center;
        }
        .buttons 
        {
            text-align: center;
        }
        .jumbotron {
            position: relative;
            background: #dca7ea url("dogGif.gif") center center;  
            width: 50%;
            display: block;
            margin-left: auto;
            margin-right: auto;
            height: 50%;
            background-repeat: no-repeat;
            overflow: hidden;
        }
        .jumbotron2 
        {
              min-height: 50%;  /* Fallback for browsers do NOT support vh unit */
              min-height: 40vh; /* These two lines are counted as one :-)       */
              background-color: #fb6f74;
              display: flex;
              align-items: center;
        }
        #dbResult
        {
            text-align: center;

        }
    </style>

    <body style = "background-color: #dca7ea;">
        <h1>Pet Name generator! The perfect name for your pet.</h1>
        <div class="jumbotron">
            <div class="container">
           </div> 
        </div>

        <div class="jumbotron2">
            <div class="container">
                  <div class="radio">
                      <label><input type="radio" name="girl">Girl</label>
                      <label><input type="radio" name="boy">Boy</label>
                      <button type="button" class="btn btn-default">Normal names</button>
                      <button type="button" class="btn btn-default">Specialty names</button> 
                      <a href="file2.php" target="_blank"><button id="submit_btn" onclick="myAjax()" type="button" class="btn btn-default">Random name</button></a> 
                      <p method="post" action="file2.php" id="dbResult" name="databaseNames">No results </p>
                  </div>

            </div>
        </div>
    </body>
</html>

My query.php file is as follows:

<?php 
    $db = mysqli_connect('localhost','root','','names_tbl');
    if($db->connect_errno > 0){
        die('Unable to connect to database [' . $db->connect_error . ']');
    }

    $query = "SELECT * FROM names_tbl.general_names LIMIT 5;";
    $result = mysqli_query($db, $query);
    while($row = mysqli_fetch_assoc($result))
    {
        echo "<u>".$row['Name']."</u>";            
    }
?>

Finally, my file I want the data to be transferred on is giving an error saying "Undefined index: dbNames" in file2.php(this is the line in which it's giving me the error in my second file)

          <?php
             echo("Names: " . $_POST['dbNames'] . "<br/>\n");
          ?>

No SO posts have been able to help me, so any help on this would be greatly appreciated.

I have already looked at these posts and nothing has worked for me. I wouldn't have posted the question if I didn't look at the popular posts with my question — beastlyCoder, Nov 05 '17 at 19:06
Do a Google Search for Ajax send variables to php script. You should find something — Rotimi, Nov 05 '17 at 19:10
You are not attaching data to your post request, `data : { }` is empty, while it should be `data : {'dbNames': 'My data I want to attach'}`. — Xorifelse, Nov 06 '17 at 23:27

Xorifelse · Accepted Answer · 2017-11-06T23:50:08.563

First of all, you should not embed an ajax request to a function. It strips off a large portion of flexibility of something called "events".

You want to request data when the user clicks on the button:

<script type = "text/javascript">
    $(function() {
      $('#normalnames').click(function(){
        $.ajax( { 
          type : 'POST',
          data : { type: 'normalnames' },
          url  : 'query.php',
          success: function ( data ) {
            $('#dbResult').html(data);
          },
          error: function ( xhr ) {
            alert( "error" );
          }
        });
      });

      $('#specialnames').click(function(){
        //etc
      });
    });
</script>

<button id="normalnames" type="button" class="btn btn-default">Normal names</button>
<button id="specialnames" type="button" class="btn btn-default">Normal names</button>

And on the PHP side of things:

<?php
  if(isset($_POST['type'])){
    if($_POST['type'] == 'normalnames'){
      // make normal names query
    } elseif($_POST['type'] == 'specialnames'){
      // make special names query
    } else {
      die('incorrect request');
    }
  } else {
    die('unknown request');
  }
?>

Now you want to enlist data in the dbresponse div:

while($row = mysqli_fetch_assoc($result)){
    $data[] = "<li>".$row['Name']."</li>";            
}

echo '<ul>' . implode($data) . '</ul>';

That would be the gist of it, perhaps its better to work with returning with json (echo json_encode($data) without the html tags) as javascript can traverse though it and append it to the list.

To use that data on another page, use sessions.

Passing data from one PHP page to another PHP Page

1 Answers1