Hello guys, I have problem with the script i was coding. This is a simple news script. I am not able to add information to the database. I don't know what Im doing wrong. Here is my php code:
<?php
if (isset($_POST['submit'])) {
$title=$_POST['title'];
$small_text =$_POST['small_text'];
$full_text=$_POST['full_text'];
include("connectmysql.php");
mysql_query("insert into main (id,title,small_text,full_text) values('$id','$title','$small_text','$full_text')");
}
?>
here is html:
<FORM name="forma2" action="?" method="post" enctype="multipart/form-data">
<TABLE>
<TR>
<TD>Title</TD>
<TD>
<input type="text" size="70" name="title">
</TD>
</TR>
<TR>
<TD>small text</TD>
<TD>
<input type="text" size="80" name="small_text">
</TD>
</TR>
<TR>
<TD>full text</TD>
<TD>
<textarea cols="60" rows="12" name="full_text"></textarea>
</TD>
</TR>
<TR>
<TD>picture</TD>
<TD><input type="file" name="picture"></TD>
</TR>
<TR>
<TD>picture</TD>
<TD><input type="file" name="picture2"></TD>
</TR>
<TR>
<TD>picture</TD>
<TD><input type="file" name="picture3"></TD>
</TR>
<TR>
<TD>picture</TD>
<TD><input type="file" name="picture4"></TD>
</TR>
<TR>
<TD>picture</TD>
<TD><input type="file" name="picture5"></TD>
</TR>
<TR>
<TD>picture</TD>
<TD><input type="file" name="picture6"></TD>
</TR>
<TR>
<TD></TD>
<TD><input type="submit" value="submit"></TD>
</TR>
</TABLE> </form>
I tried to use $id=Time("U"); and then i used $id=$_POST[id] but didn't get it working. Any help please?
Thanks in advance
