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ajax jquery php

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Kastaji Wibowo answered on October 25, 2020 Popularity 10/10 Helpfulness 2/10

answer ajax jquery php

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ajax jquery php

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<!doctype html>
<html>
<head>
<script src="https://code.jquery.com/jquery-3.3.1.js" integrity="sha256-2Kok7MbOyxpgUVvAk/HJ2jigOSYS2auK4Pfzbm7uH60=" crossorigin="anonymous"></script>
</head>
<body>
<form id="loginform" method="post">
    <div>
        Username:
        <input type="text" name="username" id="username" />
        Password:
        <input type="password" name="password" id="password" />    
        <input type="submit" name="loginBtn" id="loginBtn" value="Login" />
    </div>
</form>
<script type="text/javascript">
$(document).ready(function() {
    $('#loginform').submit(function(e) {
        e.preventDefault();
        $.ajax({
            type: "POST",
            url: 'login.php',
            data: $(this).serialize(),
            success: function(response)
            {
                var jsonData = JSON.parse(response);
  
                // user is logged in successfully in the back-end
                // let's redirect
                if (jsonData.success == "1")
                {
                    location.href = 'my_profile.php';
                }
                else
                {
                    alert('Invalid Credentials!');
                }
           }
       });
     });
});
</script>
</body>
</html>

Popularity 10/10 Helpfulness 2/10 Language whatever

Source: code.tutsplus.com

Tags: ajax jquery php whatever

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Contributed on Jun 01 2022

Kastaji Wibowo

0 Answers Avg Quality 2/10

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PHP - AJAX and PHP

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<html>
<head>
<script>
function showHint(str) {
  if (str.length == 0) {
    document.getElementById("txtHint").innerHTML = "";
    return;
  } else {
    var xmlhttp = new XMLHttpRequest();
    xmlhttp.onreadystatechange = function() {
      if (this.readyState == 4 && this.status == 200) {
        document.getElementById("txtHint").innerHTML = this.responseText;
      }
    };
    xmlhttp.open("GET", "gethint.php?q=" + str, true);
    xmlhttp.send();
  }
}
</script>
</head>
<body>

<p><b>Start typing a name in the input field below:</b></p>
<form action="">
  <label for="fname">First name:</label>
  <input type="text" id="fname" name="fname" onkeyup="showHint(this.value)">
</form>
<p>Suggestions: <span id="txtHint"></span></p>
</body>
</html>

Popularity 10/10 Helpfulness 2/10 Language php

Source: Grepper

Tags: ajax

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Contributed on May 23 2022

naly moslih

0 Answers Avg Quality 2/10

PHP - AJAX and MySQL

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<html>
<head>
<script>
function showUser(str) {
  if (str == "") {
    document.getElementById("txtHint").innerHTML = "";
    return;
  } else {
    var xmlhttp = new XMLHttpRequest();
    xmlhttp.onreadystatechange = function() {
      if (this.readyState == 4 && this.status == 200) {
        document.getElementById("txtHint").innerHTML = this.responseText;
      }
    };
    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>

Popularity 9/10 Helpfulness 2/10 Language php

Source: Grepper

Tags: ajax

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Contributed on May 23 2022

naly moslih

0 Answers Avg Quality 2/10

ajax project using php and jquery

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<?php
if (isset($_POST['username']) && $_POST['username'] && isset($_POST['password']) && $_POST['password']) {
    // do user authentication as per your requirements
    // ...
    // ...
    // based on successful authentication
    echo json_encode(array('success' => 1));
} else {
    echo json_encode(array('success' => 0));
}

Popularity 8/10 Helpfulness 2/10 Language php

Source: code.tutsplus.com

Tags: ajax jquery jqu

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Contributed on Oct 25 2020

GrepperSasi

0 Answers Avg Quality 2/10

AJAX PHP Example

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<p>Start typing a name in the input field below:</p>
<p>Suggestions: <span id="txtHint"></span></p>

<form>
First name: <input type="text" onkeyup="showHint(this.value)">
</form>

<script>
function showHint(str) {
  if (str.length == 0) {
    document.getElementById("txtHint").innerHTML = "";
    return;
  } else {
    const xmlhttp = new XMLHttpRequest();
    xmlhttp.onload = function() {
      document.getElementById("txtHint").innerHTML = this.responseText;
    }
  xmlhttp.open("GET", "gethint.php?q=" + str);
  xmlhttp.send();
  }
}
</script>

Popularity 10/10 Helpfulness 1/10 Language php

Source: Grepper

Tags: ajax

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Contributed on May 23 2022

naly moslih

0 Answers Avg Quality 2/10

ajax jquery php

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