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How to execute a PHP query to select an option using AJAX?

Well, I know that this question was answered ( Run a PHP script on the same page after selecting the dropdown options using Ajax or JavaScript ), but the answers weren’t very useful for those who have never used AJAX before. How to run a query, for example, the one that was created if someone selects a parameter from the drop-down list?

<head>
 <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
</head>

<h3>Subject</h3>
<select name="allbooks" >
              <option value="none" ></option>
              <option value="allbooks" >All Books</option>
    </select>

<?php 
$query = "SELECT * FROM books" or die("Error in the consult.." . mysqli_error($connection)); 
    $books = $connection->query($query);
?>
+4
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3 answers

First, you need to invoke an AJAX request using Javascript. But I will guide you using jQuery (Javascript library).

Your HTML:

<select name="allbooks" id="allbooks">
  <option value="none" ></option>
  <option value="allbooks" >All Books</option>
</select>
<div id="show">
  <!-- ITEMS TO BE DISPLAYED HERE -->
</div>

jQuery.

script:

<script src="jquery-1.9.1.min.js"></script> <!-- CHANGE THE JQUERY FILE DEPENDING ON THE VERSION YOU HAVE DOWNLOADED -->
<script type="text/javascript">
  $(document).ready(function(){ /* PREPARE THE SCRIPT */
    $("#allbooks").change(function(){ /* WHEN YOU CHANGE AND SELECT FROM THE SELECT FIELD */
      var allbooks = $(this).val(); /* GET THE VALUE OF THE SELECTED DATA */
      var dataString = "allbooks="+allbooks; /* STORE THAT TO A DATA STRING */

      $.ajax({ /* THEN THE AJAX CALL */
        type: "POST", /* TYPE OF METHOD TO USE TO PASS THE DATA */
        url: "get-data.php", /* PAGE WHERE WE WILL PASS THE DATA */
        data: dataString, /* THE DATA WE WILL BE PASSING */
        success: function(result){ /* GET THE TO BE RETURNED DATA */
          $("#show").html(result); /* THE RETURNED DATA WILL BE SHOWN IN THIS DIV */
        }
      });

    });
  });
</script>

get-data.php, , AJAX.

if(!empty($_POST["allbooks"])){
  /* DO YOUR QUERY HERE AND GET THE OUTPUT YOU WANT */
  echo $output; /* PRINT THE OUTPUT YOU WANT, IT WILL BE RETURNED TO THE ORIGINAL PAGE */
}

- JSfiddle.

+13
 var id="1";
 $.ajax({
    type: 'POST',
    url: 'yourphppage',
    dataType: "json",
    data: {
        idofrow:id
    },
    success: function(data) {
        alert(data);
    },
    error: function(data) {
        alert(data);
    }
});

ajax, , , , , , json text.

php-

$id = ($_POST['idofrow']);

,

SELECT * FROM table where rowid = $id

.

+1

Check out this simple tutorial. Hope this helps.

      <html>
    <head>
    <script>
    function showUser(str) {
        if (str == "") {
            document.getElementById("txtHint").innerHTML = "";
            return;
        } else { 
            if (window.XMLHttpRequest) {
                // code for IE7+, Firefox, Chrome, Opera, Safari
                xmlhttp = new XMLHttpRequest();
            } else {
                // code for IE6, IE5
                xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
            }
            xmlhttp.onreadystatechange = function() {
                if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
                    document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
                }
            }
// getuser.php is seprate php file. q is parameter 
            xmlhttp.open("GET","getuser.php?q="+str,true);
            xmlhttp.send();
        }
    }
    </script>
    </head>
    <body>

    <form>
    <select name="users" onchange="showUser(this.value)">
      <option value="">Select a person:</option>
      <option value="1">Peter Griffin</option>
      <option value="2">Lois Griffin</option>
      <option value="3">Joseph Swanson</option>
      <option value="4">Glenn Quagmire</option>
      </select>
    </form>
    <br>
    <div id="txtHint"><b>Person info will be listed here...</b></div>

    </body>
    </html>

Getuser.php file

   <!DOCTYPE html>
    <html>
    <head>
    <style>
    table {
        width: 100%;
        border-collapse: collapse;
    }

    table, td, th {
        border: 1px solid black;
        padding: 5px;
    }

    th {text-align: left;}
    </style>
    </head>
    <body>

    <?php
// don't use intval if your select value is not numberic
    $q = $_GET['q'];

    $con = mysqli_connect('localhost','peter','abc123','my_db');
    if (!$con) {
        die('Could not connect: ' . mysqli_error($con));
    }

    mysqli_select_db($con,"ajax_demo");
    $sql="SELECT * FROM user WHERE id = '".$q."'";
    $result = mysqli_query($con,$sql);

    echo "<table>
    <tr>
    <th>Firstname</th>
    <th>Lastname</th>
    <th>Age</th>
    <th>Hometown</th>
    <th>Job</th>
    </tr>";
    while($row = mysqli_fetch_array($result)) {
        echo "<tr>";
        echo "<td>" . $row['FirstName'] . "</td>";
        echo "<td>" . $row['LastName'] . "</td>";
        echo "<td>" . $row['Age'] . "</td>";
        echo "<td>" . $row['Hometown'] . "</td>";
        echo "<td>" . $row['Job'] . "</td>";
        echo "</tr>";
    }
    echo "</table>";
    mysqli_close($con);
    ?>
    </body>
    </html>
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Source: https://habr.com/ru/post/1605872/

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