[html php form problem]

use quotes around option value or use select box as

<select name="charity_name" id="charity_name">
<?php
    while ($row=mysql_fetch_array($charity))
    {
       echo '<option value="'.$row[charity_name].'">'.$row[charity_name].'</option>'; 
    }
?>
</select>

hope this will help you

[PHP with MYSQL]

That solution isn't going to quite work since $row{"visivel"} comes from the first query.

 

 

@magcr23 - That first query probably is not necessary. The update query will only change visivel when it is equal to zero.

 

If you need to show the message about the product already being shown, you could use mysql_affected_rows(). More information about the function can be found here:

http://php.net/manual/en/function.mysql-affected-rows.php

 

oh sorry, i missed that

[PHP with MYSQL]

You missed mysql_query() for UPDATE query, also SELECT query should be below UPDATE query, so you will get updated data

<?php
	require("bd/conexao.php");       
	if ($row{"visivel"}== 0)
	{
		$update= "UPDATE produtos SET visivel = 1 WHERE visivel = 0";
		mysql_query($update);
		echo $update;
	}
	else
	{
		echo "o produto ja esta visivel";
	}   

	$sql = "SELECT * FROM produtos";
	$resultado = mysql_query($sql);
	$row = mysql_fetch_array($resultado);
?>

[get total post count]

Try this query

SELECT username, count(username) AS post_count FROM TABLE_NAME 
GROUP BY username
ORDER BY count(username) DESC;

you can apply LIMIT to get top records, as query will give sorted result according to max post count of user.

[mysqli_fetch_array() problem]

you need to use

$result = mysqli_query($link, $query);

before

$result_array = mysqli_fetch_array($result);

or check this URL: http://php.net/manual/en/mysqli-result.fetch-array.php

[Increase import limit from 16384 Kib]

have you restarted the XAMPP after updating the php.ini file?

 

if yes, then it should work

[Problem showing top 10 highest score]

SELECT TOP is not supported in MySQL, you need to use LIMIT

 

try this

SELECT ID, Name, Score FROM Scores ORDER BY Score DESC LIMIT 10

[Page Pagination (php/mysql)]

Check this script online

 

http://www.php-guru.in/2014/pagination-php-mysql/

[Deprecated: mysql_connect()]

mysql_connect() function is not supported in PHP 5.5.0 and your PHP version may have been upgraded to 5.5.0

 

check this url: http://php.net/manual/en/function.mysql-connect.php

[What am I doing wrong?]

echo tha $_POST['who] value at the top and check what value you are getting after submitting the form.

[Alert message when quantity reaches max or min number]

Yes, use AJAX to get current quantity($qty) value

[get month value from date stamp]

Try this:

echo date("m",strtotime('2014-08-20 07:32:20'));

[pdf report]

Try TCPDF

[PHP WSDL Help Needed]

for PHP SOAP you can try NuSOAP Toolkit

[Noob Q: Simple JS Alert POST callback to php script]

you can use AJAX 

[how to highlite current date in calendar script]

you can check it in jsfiddle for the output

[how to highlite current date in calendar script]

search the following line 

html += '<td>' + i + '</td>';

and replace it with

if (i == d.getDate())
    html += '<td style="color:blue;">' + i + '</td>';
else
    html += '<td>' + i + '</td>';

[Getting to a value from an Array]

Use

$result = json_decode($string,true);

you will get $result in array format

[Displaying Even and Odd SQL Results]

Check this simple logic

 

<table border="0" cellspacing="0" cellpadding="5">
<tr>
<?php
    $array = array(1,2,3,4,5,6);
    $i = 1;
    foreach($array as $val)
    {
        echo '<td>'.$val.'</td>';
        if ($i % 2 == 0)
             echo "</td></tr>";
      $i++;
    }
?>
</tr>
</table>

I hope this will help you

[Parse error: syntax error, unexpected T_ENDWHILE]

you have only 1 while loop and 2 endwhile

[QUICK help with making a variable a link]

You can use it like this also

<?php
     $link = '<a href="post?id='.$id.'&title='.$slug_title.'">'.$title.'</a>';
?>

[php mysql simple script not working.]

you need to check the MySQL query

$query = urldecode($query);

echo the above line and check the query

 

 

[Does anyone know of a WYSIWYG HTML editor]

try FCKeditor or TinyMCE

[Split a form on the same page]

why do you want to split the form, it can be done in single form

[MySql insert error]

you have used "Like" as column name, it is reserved word in mysql