Foreach loop and variables in php

Question

Happy new yar. I am still new to php so I need your great help. I am designing a post page where visitors can post anything. I am almost through but when I post something to the page, the new post over ride the old one, I am certain I need to use foreach loop but my problem I can't define the $posting variable.

My bad, but I really need you guys to help me. here is my coding:

<?php
include 'connect.php';
?>

<?php
if (isset($_POST['submit'])) {
    $title = $_POST['title'];
    $post = $_POST['post'];
    $name = $_POST['name'];

    if (empty($title) or empty($post) or empty($name)) {
        $message = "Please fill in all fields";
    } else {
        mysql_query("INSERT INTO prayer VALUES('', '" . $title . "', '" . $post . "', '" . $name . "')");
        $message = "PRAYER REQUEST SUBMITTED!!!";
    }

    echo"$message";
}
?>
<form method="post" action="prayerpage.php">
    <table width="80%">
        <tr>
            <td><b>Name:</b><input type="text" name="name" />
                <b>Title:</b><input type="text" name="title" /></td>
        </tr>
        <tr>
            <td><b>Prayer<br>Request:</b></td>
            <td><textarea name='post' rows='10' cols='40'></textarea></td>
        </tr>
        <tr>
            <td></td>
            <td><input type="submit" name="submit" value="SUMIT"/></td>
        </tr>
    </table>
</form>
<hr width="70%">
<?php

function postid($id) {
    $array = array();
    $q = mysql_query("SLECT * FROM prayer WHERE id='.$id.'");
    while ($r = mysql_fetch_assoc($q)) {
        $array['id'] = $r['id'];
        $array['title'] = $r['title'];
        $array['name'] = $r['name'];
        $array['post'] = $r['post'];
    }
    return $array;
}

foreach ($posting as $posting) {
    ?>
    <table width="60%">
        <tr>
            <td><font color="blue"><?php echo $title; ?></font></td>
        </tr>
        <tr>
            <td><?php echo $post; ?> - <font color="blue"><?php echo $name; ?></font>         </td>
        </tr>
    </table>
    <hr noshade width="50%">
    <?php
}
?>

And please i also need the code to make the post a link to its page

Answer 1

check your id is auto increment or not if not then make it autoincrement

then try to change this code in your page :

mysql_query("INSERT INTO prayer VALUES('', '".$title."', '".$post."', '".$name."')");

to

mysql_query("INSERT INTO prayer VALUES('".$title."', '".$post."', '".$name."')");

Answer 2

There are a couple problems with your code. The first is that you mysql_query when you should be using mysqli_query, same mysql_fetch_assoc, should be mysqli_fetch_assoc -- please replace all references of mysql_* with mysqli_*. The second is you are running a query inside of a function. The third problem is your query is wrong. And... you do not need a foreach in the way you are using it.

function postid($id) {
    global $db;
    $array = array();
    $id = (int)$id; // cast as int to prevent SQL injection
    $q = mysqli_query($db, "SELECT * FROM prayer WHERE id=$id");
    while ($r = mysqli_fetch_assoc($q)) {
        $array['id'] = $r['id'];
        $array['title'] = $r['title'];
        $array['name'] = $r['name'];
        $array['post'] = $r['post'];
    }
    return $array;
}

You had a typo, it should be SELECT not SLECT . You do need to concatenate variables when you are inside of a double quoted string. I would link to the PHP documentation about it but is not super clear. Basically the following lines all result in the same output:

$amazing = "neato!";
$example1 = "This is my variable: $amazing"; // Output: This is my variable: neato!
$example2 = 'This is my variable: ' . $amazing;  // Output: This is my variable: neato!
$example3 = "This is my variable: " . $amazing;  // Output: This is my variable: neato!

Notice how you can put a variable inside of a string with double quotes. But you cannot do this:

$doesNotWork = 'This is my variable: $amazing'; // Output: This is my variable: $amazing
$doesNotWork = "This is my variable: '.$amazing.'"; // Output: This is my variable: '.$amazing.'

The reason you should use mysqli_query is because mysql_query is no longer supported. Inside of your function you will need to use the special keyword global to get the variable of your database connection (I set it to $db in the example but you might have it called something else).

Basically, global is a PHP keyword that will allow you to access variables that have been defined in the global scope. Read about in the documentation by clicking here

[..] within user-defined functions a local function scope is introduced. Any variable used inside a function is by default limited to the local function scope.

Lastly.. you do not have the variable $posting defined correctly.

You can fix this by getting rid of your call to foreach, replace this line:

foreach ($posting as $posting) {

With these 2 lines:

$q = mysqli_query($db, "SELECT * FROM prayer");
while ($posting = mysqli_fetch_assoc($q)) {

Your $posting variable is undefined, when it looks like you actually just want to query the database and get all the rows from the prayer table.

Answer 3

This query will give you warnings in MYSQL but execute due to PK auto increment:

mysql_query("INSERT INTO prayer 
            VALUES('', '" . $title . "', '" . $post . "', '" . $name . "')");

Solution is that define column names in this query and if your id is PK than not use because its auto increment:

mysql_query("INSERT INTO prayer (title,post,name) 
             VALUES('".$title."','".$post."','".$name."')");