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Make an AJAX load for content
Basic form validationPopulating form contents from GET - especially radio buttonsManaging network address informationBasic PHP comment formInterupt redirect to catch potential newsletter signupClass to manage config data to allow DILoad page content on hashchangeAjax call with Spring MVC ControllerUpdate game content via AJAXusing $_POST array to prepare PDO statement with variables
$begingroup$
I have code that I want to make more load of the content:
<div class="single-widget single-post">
<?php
include "config/db.php";
$query = mysqli_query($con,"SELECT * FROM posts ORDER BY id_posts DESC");
?>
<form action="" method="post">
<?php if(mysqli_num_rows($query)>0) ?>
<?php
$no = 1;
while($data = mysqli_fetch_array($query))
?>
<input type="text" name="id_posts" value="<?php echo $data['id_posts']; ?>">
<input type="text" name="comment" placeholder="Give your comment here" style="width: 100%;" required>
<input type="submit" name="submit" value="submit" />
<?php $no++; ?>
<?php ?>
</form>
</div>
Please give me a detailed explanation. If you do, give me easy code for making,
javascript php jquery ajax
New contributor
$endgroup$
add a comment |
$begingroup$
I have code that I want to make more load of the content:
<div class="single-widget single-post">
<?php
include "config/db.php";
$query = mysqli_query($con,"SELECT * FROM posts ORDER BY id_posts DESC");
?>
<form action="" method="post">
<?php if(mysqli_num_rows($query)>0) ?>
<?php
$no = 1;
while($data = mysqli_fetch_array($query))
?>
<input type="text" name="id_posts" value="<?php echo $data['id_posts']; ?>">
<input type="text" name="comment" placeholder="Give your comment here" style="width: 100%;" required>
<input type="submit" name="submit" value="submit" />
<?php $no++; ?>
<?php ?>
</form>
</div>
Please give me a detailed explanation. If you do, give me easy code for making,
javascript php jquery ajax
New contributor
$endgroup$
add a comment |
$begingroup$
I have code that I want to make more load of the content:
<div class="single-widget single-post">
<?php
include "config/db.php";
$query = mysqli_query($con,"SELECT * FROM posts ORDER BY id_posts DESC");
?>
<form action="" method="post">
<?php if(mysqli_num_rows($query)>0) ?>
<?php
$no = 1;
while($data = mysqli_fetch_array($query))
?>
<input type="text" name="id_posts" value="<?php echo $data['id_posts']; ?>">
<input type="text" name="comment" placeholder="Give your comment here" style="width: 100%;" required>
<input type="submit" name="submit" value="submit" />
<?php $no++; ?>
<?php ?>
</form>
</div>
Please give me a detailed explanation. If you do, give me easy code for making,
javascript php jquery ajax
New contributor
$endgroup$
I have code that I want to make more load of the content:
<div class="single-widget single-post">
<?php
include "config/db.php";
$query = mysqli_query($con,"SELECT * FROM posts ORDER BY id_posts DESC");
?>
<form action="" method="post">
<?php if(mysqli_num_rows($query)>0) ?>
<?php
$no = 1;
while($data = mysqli_fetch_array($query))
?>
<input type="text" name="id_posts" value="<?php echo $data['id_posts']; ?>">
<input type="text" name="comment" placeholder="Give your comment here" style="width: 100%;" required>
<input type="submit" name="submit" value="submit" />
<?php $no++; ?>
<?php ?>
</form>
</div>
Please give me a detailed explanation. If you do, give me easy code for making,
javascript php jquery ajax
javascript php jquery ajax
New contributor
New contributor
edited 11 mins ago
Jamal♦
30.4k11121227
30.4k11121227
New contributor
asked 25 mins ago
Febry AryoFebry Aryo
1
1
New contributor
New contributor
add a comment |
add a comment |
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Febry Aryo is a new contributor. Be nice, and check out our Code of Conduct.
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Febry Aryo is a new contributor. Be nice, and check out our Code of Conduct.
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