[send variables]

please stop bumping your thread(s)..and no your question doesnt make sense to me

[Selecting number from a TXT file]

the leading zeroes could cause an issue... try trimming your values

$var = ltrim($var, '0'); 

[syntax for picking mysql column from form]

what does your mysql table look like?

[not uploading]

i notice that in this line

$uploadfile = $target_path . basename($_FILES[uploadedfile][name][$key]);

your forget to quote your indices and should be

$uploadfile = $target_path . basename($_FILES['uploadedfile']['name'][$key]);

also, you assign a value to $id twice...so your $_POST['id'] is being overwritten

[not uploading]

well you definitely over complicated the upload process..try to shorten your code and decrease the amount of possible things that can go wrong

[What's wrong with my query...]

how are you displaying the data?

[sort an array but ignore a part of the values (better explained in the post)]

i still need the numbers at the end of the strings/values then on the end when echoing it. i just need the sort(); function to ignore them (the numbers at the end behind -aa-zz-) when sorting. the sort(); should sort only till aa-zz- and not by the number at the end.

 

for example, this doesn't work the right way now:

 

<?php
$a=array(
'222-aa-zz-6',
'333-aa-zz-5',
'333-aa-zz-4'
);
sort($a);
foreach($a as $key=>$values){

$strpos=strpos($values, "z-");
$values1=substr($values,0,$strpos);
echo $values.' ('.$values1.')<br />';
}
?>

 

as it echoes this:

 

222-aa-zz-6 (222-aa-z)

333-aa-zz-4 (333-aa-z)

333-aa-zz-5 (333-aa-z)

 

instead of this:

 

222-aa-zz-6 (222-aa-z)

333-aa-zz-5 (333-aa-z)

333-aa-zz-4 (333-aa-z)

thought thats what you wanted...and abra is right... the numbers on the end are irrelevant when using sort()

[New here]

Yup, thanks

?

[Resizing Uploaded Images]

also, it doesnt upload to you img dir because you stated that it should go to the dir above that here

define('UPLOAD_DIR', 'C:/upload_test/');

[PHP Count down and Time Zones]

Check this out!

 

http://php.net/manual/en/function.date-default-timezone-set.php

same link i posted...lol...just realized my sentence got all jumbled..weird

[PHP Count down and Time Zones]

you can set the default timezone ofhere the script by using the function

[UPDATE of database table]

what exactly is your issue? and mysql_close() will close the connection to the mysql server that is associated with the link identifier...but is not actually necessary here because the connection will automatically close when the script is finished executing...

[not uploading]

same thing happened, no errors came up though  :( :( :confused:

well if you didnt receive the error in the else statement then your file must have been moved..perhaps if you add some more smileys to your posts your question might be resolved...

[Auto populate and Post with mysql]

but what im asking is...if you arent able to receive a value when you echo $row['city_name'], then how is it that $row['city_name'] has a value in your option tag..

[parse error]

fix your problem?

[parse error]

this

$query = "INSERT INTO contact_us (rate_us,your_name,email,your_comments)"."VALUES ('$rate_us','$your_name','$email','$your_comments');

should be

$query = "INSERT INTO contact_us (rate_us,your_name,email,your_comments) VALUES ('$rate_us','$your_name','$email','$your_comments')";

[Auto populate and Post with mysql]

if nothing works than how is your option tag being populated with your city_name?

echo "<option value='{$row['city_code']}'>{$row['city_name']}</option>";

how would i go POST the city_name when pushing submit, if it's even possible?

what exactly are you trying to ask?

[not uploading]

have you tried inserting some code to debug?

if (move_uploaded_file($_FILES['uploadedfile']['tmp_name'][$key], $uploadfile)) { echo $value . ' uploaded<br>'; }else {echo "error";}

etc...

[not uploading]

no errors are triggered?

[Auto populate and Post with mysql]

its blank because you are trying to echo it outside of your while loop ....try

echo "<form method='POST' action =''>";
echo "<select name='cityd'>";
while($row = mysql_fetch_array($res))
{
echo "<option value='{$row['city_code']}'>{$row['city_name']}</option>";
}
echo "</select>";
echo "<input type='submit' name='weaCode' value='Submit'/>";
echo "</form>";
echo $_POST["cityd"];
while($row = mysql_fetch_array($res))
{
echo $row['city_name'];
}
?>

[PHP in Ajax Call Fails]

sorry for being unclear..i'm talking about $gid and $status

[Forgot password, php.]

your host var must be wrong

[Auto populate and Post with mysql]

what do you receive when you echo $row['city_name']?

[PHP in Ajax Call Fails]

where do you assign your $_GET values to variables?

[sort an array but ignore a part of the values (better explained in the post)]

sort($a);
foreach($a as $key=>$values) {

   $strpos = strpos($values, "z-");
   $values = substr($values,0,$strpos);
   echo $values;
}