j5646w

I’m currently writing a website in Windows XP with Apache 2.2, mySQL and php. I would like to know if there is a way for the website to automatically receive emails and put it into mysql . Other words the website must check for mails every so often and then insert the sender, subject and the mail into a database and if possible to put in rules as to the subject?

Good day, I new to PHP I am having problems with a two dropdowns on a form, can someone please tell me where I'm going wrong. [attachment deleted by admin]

Hi there I have a problem with the following code If I select the dropdown option it shows - ad.php?id=animal& region= The Value is not been retrieved - what am I doing wrong? <select onChange="window.location='ad.php?id=<? print ($type); ?>& region=' + this.value" id="reg" name="reg" title="Select Region"> <option value"">All Region</option> <option value"eastern cape">Eastern Cape</option> <option value"free state">Free State</option> <option value"gauteng">Gauteng</option> <option value"kwazulu natal">KwaZulu Natal</option> <option value"limpopo">Limpopo</option> <option value"mpumalanga">Mpumalanga</option> <option value"northern province">Northern Province</option> <option value"western cape">Western Cape</option> </select> </form>

Everything Do I use <a href="/petonline.co.za/data/imagedisplay.php?"$user_uid"="$uidadd"">More Info</a> - but this doesnt work? I tried to google it but I can find a manual for dummies!

Hi there (I'm new at PHP/mysql) I am busy with a project - I maneged to load info into the mysql database and display the results BUT - now I need some Help I can display the result - now I want to have the ability to click on the result and it must open a new page with only the selected result information?

Hi ther ManiacDan, thanks for the reply I tried the JOIN syntax But if I echo now it shows the results twice? $result = mysql_query("SELECT * FROM other INNER JOIN images ON other.other_id=images.id WHERE other_note = 'Lost' ORDER BY images.date DESC limit 20");

Okay I have 3 tables 1) Images - "image_id" "image_url" "image_thumb" "image_date" 2) User - "user_id" "user_name" "user_surname" 3) Other - "other_id" "other_note" The image_id , user_id, other_id have the same number (to link them) my question is how will I go about linking the three and echoing the information. Keep in mind that some users don't have images <?php include('config.php'); $result = mysql_query("SELECT * FROM other, images, user WHERE other.other_uid = 1 AND images.image_id = 1 AND user.user_uid = 1 ORDER BY images.date DESC limit 5"); $url='http://localhost/petonline.co.za/data/'; while($row = mysql_fetch_array($result)){ echo "<B>".$row['pet_depo']."</b><br/>, Photo: <img src='".$url.$row['thumb']."'>, Name: ".$row['user_uid']."<br/><br/><br/>"; }