you can. if you use the following code this will work if your server will allow you to use .htaccess files.

php_value post_max_size 128M
php_value upload_max_filesize 128M
php_value...

is it possible to change the max file upload and the memory limit allowed for file uploads in php with a .htaccess file.

the server my website is held on will not give me access to the php.ini...

thanks once again for your help.

i had tried nearly every variation of putting the echo in different places and closing it out on different lines. just didnt think about the fact that the page...

this code was working fine until i changed the way the site loads the content (it now loads the content into a div tag) and since then it will not populate the menu.

the page can be viewed at
...

very strange i nearly pulled all my hair out yesterday trying to fix that problem. no worries i got a new problem now i will start a new post.

i removed the centre from the table and it worked fine but before that the scrollbar just wouldnt show. i was viewing it first in ie7 with the table centred and the scrollbar was not visible.

sorted. took the centre from the table in the second section of the code and it has decided to work now.

if someone could shed any light on why this happened please do so.

i have these two pages of information which are stored on a php page which then feeds the main div tag on the index page.

the problem i have here is that a scrollbar shows up on the first page but...

it was a spelling mistake stopping

i have built a page which loads content into a div tag from a php page using javascript.

the problem i am having is that the title for eachg page is not inside the div so i keep being left with...

if you look for the section where the values of $consname and $consname2 are set you can change where the files are stored on the server there.
if you want the oiginal image stored in the images...

thank you for the reply i ended up getting it working late last night with a similar piece of code. thank you again for the reply.

a foreach loop will need to start with a while loop and i do not see the need to use the id of the table as i only need to loop through the email address and user these to send the email.

i am trying to get this code to work but there is something wrong with the array (not to good at these). the main part of the code works it is sending out the email but only ever to one registered...

this code stores the path to the image in the db an stores two copies of the image on the server. the original and the newly created thumbnail image.

you will need to create a folder called image...

the sleep function pauses the mysql statement. i cannot see how that would solve the problem of my array not being populated.

i have changed the code slightly now and have got it sending out one email. i have narrowed the problem down to the mysql query as this is not populating the array it is only putting on email...

i am trying to write a script that will send out an email to all the email addresses stored in a mysql db when the site has been updated. the code i have so far is as follows

$sql= "INSERT INTO...

well i still got my job so i cannot be doing that bad. thanks again take it easy and have fun.

thanks for the reply nav33n i have been looking for missing semo colons. which is usually my problem. hope u all good ne way not been on here for a while. must say something about the help you...

i have this line of code which i cannot see a problem with but i keep getting an error.

$sql= "INSERT INTO emails (Firstname, Lastname, email, table) VALUES ('$a', '$b', '$c', '$table')";

the...

sorted changed the code to this and it works fine now
print "<h4>There are $num_rows records.<P></h4>";
echo "<div style=\"height: 150px; width: 100%; overflow:auto;\">";
print "<table width=600...

not sure if this is the correct place to post but here we go anyway.

what i am trying to do is put a table inside a div tag held within a php page but it is not working. here is the code

print...

the variable echo's out until the code is ran for the upload of the image then it decides to go missing and i cannot work out why this is happening.

try this code out i have no problems with the image quality with it.

<?php



set_time_limit(0);

$link = mysql_connect(localhost, xxxxxxxx, xxxxxxxxx) or die("Could not connect to...

is that some javascript?

i have a column named id which is set to auto increment. this code should update the table depending on what id number the user chose on a previous page. the id the user chose is posted over and...

as for doing this without the click of a submit but you would have to look at HTTP_REQUEST to upload the images without the click of a submit button. the code for the upload will keep the user on...

here is the code to delete all file from the server once you no they are not needed. be careful when this code is used as it will delete every file from within the specified filder in the $dir...

here is the code for the upload of the image. your mysql table should have two columns one which is called id which is set to auto increment and the other which is called image.

<?php


...

i would store the image on to the server and save the path to the image into a mysql db. once you have the path to the image inside the db you can use the mysql function to fined the last entry...

so what you want to do is upload an image on to the server so it can be used where it is needed. then if another image is uploaded you want this to be over write the old image and for the new image...

hope someone can help me with this.

i am trying to update records in a mysql db with a new image but i keep gettin errors.

i am using a hidden form input to post a variable across to a new...

i hate programming!!

not really it wouldnt work coz i aint put ' ' around my variable names in the update code.

the way it is working (i no it is not the best way to do it) is when the user uploads an image it will be to go with the last article added so th highest number in the id column will work fine in...

that section of code is working now it is giving me the number of the last entered row.

my update query is not working tho it is not putting the image into the db. here is my code for this...

spelling mistake ignor last post

it dont like this line

$lastid = mysql_fetch_row($run);

i get this error

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in...

so if i use the select max(id) would i beable to do something like

select max(id) from interal;
$lastid=max(id);
$sql = update internal set image = $image where id = $lastid;

i have set up a mysql db for storing news articles to be displayed within a flash website. i have now been asked if i could add an image upload function to this section.

i have created the upload...