toxictoad Posted December 5, 2008 Share Posted December 5, 2008 Hey I'm back again... I've got a HTML form that posts the info entered to a php page that then updates the database. I have all this working for one database so adapted it for another but for some reason it's only updating 2 fields in the database. I've gone over it repeatidly and just can't understand why it's not updating all the Fields so an extra pair of eyes to look over what I've got might help. DB Name: rss Table Name: rssfeeds Fields: NewsID - int(11) | Title - varchar(250) | Link - varchar(250) | PubDate - date | Description - varchar(250) | Content - text | Source - varchar(250) The HTML form <html> <head> <title>Add News Artcle</title> </head> <body> <form method="post" action="update.php"> Title: <input type="text" name="title" size="70" /><br /> Link: <input type="text" name="link" size="60" value="http://www.mysite.co.uk/details.php?id=" /><br /> Date: <input type="text" name="date" size="20" /><br /> Description:<br /> <textarea name="description" cols="50" rows="5"></textarea><br /> Content:<br /> <textarea name="content" cols="50" rows="12"></textarea><br /> Source: <input type="text" name="source" size="40" value="http://" /><br /> <input type="submit" value="Add Article" /><br /> </body> </html> The Update php code <?php // initiate DB connection mysql_connect ("localhost", "USERNAME", "PASSWORD") or die ('error: ' .mysql_error()); mysql_select_db ("DBNAME"); $title = $_POST['Title']; $link = $_POST['Link']; $date = $_POST['PubDate']; $description = $_POST['Description']; $content = $_POST['Content']; //$link = $_POST['link']; $source = $_POST['source']; mysql_connect ("localhost", "USERNAME", "PASSWORD") or die ('error: ' .mysql_error()); mysql_select_db ("phatjoin_rss"); // pull the id for the last article entered into the rssfeeds table $lastArticle = mysql_fetch_assoc(mysql_query("SELECT NewsID FROM rssfeeds ORDER BY NewsID DESC LIMIT 1")); $link = "http://www.mysite.co.uk/rss/details.php?id=" . ($lastArticle['NewsID'] + 1); // append the last id plus one to the link before inserting... $query="INSERT INTO rssfeeds (title, link, pubdate, description, content, source) VALUES ('".$title."', '".$link."', '".$date."', '".$description."', '".$content."', '".$source."')"; mysql_query($query) or die ('Error updating database'); echo "" ; ?> <html> <head> <title>Updated Database</title> </head> <body> Database Updated<br /> <a href="http://www.mysite.co.uk/rss/form.php">Enter New Article</a></body> </html> When I check what's been entered into the DB it's only the following NewsID | Link | Source | The Date field just shows 0000-00-00 after entering 2008-12-05 Can you see what I'm missing? Thanks Quote Link to comment Share on other sites More sharing options...
Brian W Posted December 5, 2008 Share Posted December 5, 2008 simple... name of field and you post values are different. check capitals example: $title = $_POST['Title'];// name of field is name="title" $title = $_POST['title'];//so the post variable needs to match fix them and then try again Quote Link to comment Share on other sites More sharing options...
toxictoad Posted December 5, 2008 Author Share Posted December 5, 2008 simple... name of field and you post values are different. check capitals example: $title = $_POST['Title'];// name of field is name="title" $title = $_POST['title'];//so the post variable needs to match fix them and then try again Easy when you know how I didn't realise that caps made a difference (til now) I do have a problem with the date though because it's still only showing 0000-00-00 Is there a specific way the date needs to be entered into the form? I'm entering it the way it's shown in mySQL 2008-12-05 thanks Quote Link to comment Share on other sites More sharing options...
Brian W Posted December 5, 2008 Share Posted December 5, 2008 sorry, didn't mean easy and you suck... i meant easy to fix. I made the same mistake back when i first started, till do accidentally. as for date, can't quite help you there since by the sounds of it "2008-12-05" didn't work and I don't know why it doesn't. must be a special syntax you must use which i'm not aware of. still a noob in many ways. Quote Link to comment Share on other sites More sharing options...
PFMaBiSmAd Posted December 5, 2008 Share Posted December 5, 2008 You just had one form field name that did not match the name used in the $_POST variable, would it not be prudent to check if the form field name used for your date matched? Quote Link to comment Share on other sites More sharing options...
Brian W Posted December 5, 2008 Share Posted December 5, 2008 post code again... that might help. Quote Link to comment Share on other sites More sharing options...
toxictoad Posted December 5, 2008 Author Share Posted December 5, 2008 You just had one form field name that did not match the name used in the $_POST variable, would it not be prudent to check if the form field name used for your date matched? Absolutely! I changed every reference so that they all start with capitals but still the date in the database shows as 0000-00-00 <html> <head> <title>Add News Artcle</title> </head> <body> <form method="post" action="update.php"> Title: <input type="text" name="Title" size="70" /><br /> Link: <input type="text" name="Link" size="60" value="http://www.mysite.co.uk/details.php?id=" /><br /> Date: <input type="text" name="Date" size="20" /><br /> Description:<br /> <textarea name="Description" cols="50" rows="5"></textarea><br /> Content:<br /> <textarea name="Content" cols="50" rows="12"></textarea><br /> Source: <input type="text" name="Source" size="40" value="http://" /><br /> <input type="submit" value="Add Article" /><br /> </body> </html> <?php // initiate DB connection mysql_connect ("localhost", "USERNAME", "PASSWORD") or die ('error: ' .mysql_error()); mysql_select_db ("phatjoin_rss"); $Title = $_POST['Title']; $Link = $_POST['Link']; $Date = $_POST['PubDate']; $Description = $_POST['Description']; $Content = $_POST['Content']; //$Link = $_POST['Link']; $Source = $_POST['Source']; mysql_connect ("localhost", "USERNAME", "PASSWORD") or die ('error: ' .mysql_error()); mysql_select_db ("phatjoin_rss"); // pull the id for the last article entered into the rssfeeds table $lastArticle = mysql_fetch_assoc(mysql_query("SELECT NewsID FROM rssfeeds ORDER BY NewsID DESC LIMIT 1")); $Link = "http://www.mysite.co.uk/rss/details.php?id=" . ($lastArticle['NewsID'] + 1); // append the last id plus one to the link before inserting... $query="INSERT INTO rssfeeds (Title, Link, PubDate, Description, Content, Source) VALUES ('".$Title."', '".$Link."', '".$Date."', '".$Description."', '".$Content."', '".$Source."')"; mysql_query($query) or die ('Error updating database'); echo "" ; ?> <html> <head> <title>Updated Database</title> </head> <body> Database Updated<br /> <a href="http://www.mysite.co.uk/rss/form.php">Enter New Article</a></body> </html> Quote Link to comment Share on other sites More sharing options...
PFMaBiSmAd Posted December 5, 2008 Share Posted December 5, 2008 Date (your form field name) is not equal to PubDate (your $_POST variable name.) Quote Link to comment Share on other sites More sharing options...
chronister Posted December 5, 2008 Share Posted December 5, 2008 A good tip for debugging is to echo out your query and see if it has all the data you want it to have. A side note, here is the way I write queries to avoid escaping data and such.. <?php $query="INSERT INTO rssfeeds (title, link, pubdate, description, content, source) VALUES ('$title', '$link', '$date', '$description', '$content', '$source')"; echo $query; ?> Since the whole line is wrapped with " and ", the single quotes inside don't turn the variables into literal strings like it would if you used single quotes to wrap the query. Nate Quote Link to comment Share on other sites More sharing options...
bgadberry Posted December 5, 2008 Share Posted December 5, 2008 And what would happen if someone entered data with a " in it? would that not escape out of it? I think it would be worth it to still either mysql_real_escape_string or addslashes to the data still. You really cant trust people to put in the correct information, either by mistake or by malice, someone might try to take advantage of the situation or at the very least, break it. Quote Link to comment Share on other sites More sharing options...
toxictoad Posted December 5, 2008 Author Share Posted December 5, 2008 Date (your form field name) is not equal to PubDate (your $_POST variable name.) You got it! man am I cut out for this programming stuff! (please don't answer) Thank you for your help Thanks for this chronister. I have gone through some tutorials but there's so much and I guess I'm ju7mping in the deep end so to speak without a proper understanding. Your help's much appreciated guys, thank you! Quote Link to comment Share on other sites More sharing options...
chronister Posted December 6, 2008 Share Posted December 6, 2008 Thanks for this chronister. I have gone through some tutorials but there's so much and I guess I'm ju7mping in the deep end so to speak without a proper understanding Well, learning little by little is they way to go. You are not going to get a proper understanding without making mistakes and learning from them. That is what I have done over the last few years, though I still don't have a "proper" understanding. I am going to be learning better ways to do things for a LONG TIME. This forum is filled with VERY knowledgeable people who will help you if your willing to help yourself. If you want to learn, then keep trying and when you get stuck come back here and someone will help ya. Hapy Coding! Nate Quote Link to comment Share on other sites More sharing options...
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