form to db problems

@dvdd127Nov 23.2007

[CODE]<?

mysql_connect(“localhost”, “chale”, “test”) or die(mysql_error()); mysql_select_db(“site”) or die(mysql_error());

mysql_query(“INSERT INTO ‘menu’ (‘category’,’item’,’itemdesc’,’price’,’lunch’,’dinner’) VALUES (‘$category’,’$item’,’$itemdesc’,’$price’,’$lunch’,’$dinner’)”); $category = $_POST[‘category’]; $item = $_POST[‘item’]; $itemdesc = $_POST[‘itemdesc’]; $price = $_POST[‘price’]; $lunch = $_POST[‘lunch’]; $dinner = $_POST[‘dinner’];

$note = “<DIV>The item has been added as stated below to the menu.<br>If there is a mistake you can edit it by using the search links at the top of the page for each section of the menu.</DIV>”;

print “<!– ” . $_POST[‘formaction’] . ” –> <HTML> <HEAD> <TITLE>Adding menu item</TITLE> </HEAD> <BODY>” . $note . ” <br> Item: ” . $item . ” <br> Item description: ” . $itemdesc . ” <br> Lunch item: ” . $lunch . ” <br> Dinner item: ” . $dinner . ” <br> Price: ” . $price . ” <br> </BODY> </HTML>”;

?>[/CODE]

Above is the php code I am using for submitting my form to a database… i am using form action php_self, since i am doing it all within one file. when I submit it prints out the info that was added on the screen but does not submit it to the DB…. any ideas why?

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PHP

16 Comments(s) _↴

@salmanshafiqNov 23.2007 — #check there is any error or not? by replace this line you know is there any error in the query or not

[code=php]
 
 mysql_query("INSERT INTO 'menu' ('category','item','itemdesc','price','lunch','dinner') VALUES ('$category','$item','$itemdesc','$price','$lunch','$dinner')") or die(mysql_error()); 
 [/code]

@roscorNov 23.2007 — #Try a print_r($_POST) to see what values are set.

@gpm1982Nov 23.2007 — #Move the mysql_query() line down (after $dinner = $_POST['dinner']). The code should be as follows:

[code=php]mysql_connect("localhost", "chale", "test") or die(mysql_error());
 mysql_select_db("site") or  die(mysql_error());
 
 $category = $_POST['category'];
 $item = $_POST['item'];
 $itemdesc = $_POST['itemdesc'];
 $price = $_POST['price'];
 $lunch = $_POST['lunch'];
 $dinner = $_POST['dinner'];
 
 mysql_query("INSERT INTO 'menu' ('category','item','itemdesc','price','lunch','dinner') VALUES ('$category','$item','$itemdesc','$price','$lunch','$dinner')");
 
 $note = "<DIV>The item has been added as stated below to the menu.<br>If there is a mistake you can edit it by using the search links at the top of the page for each section of the menu.</DIV>";
 
 print "<!-- " . $_POST['formaction'] . " -->
 <HTML>
 <HEAD>
 <TITLE>Adding menu item</TITLE>
 </HEAD>
 <BODY>" . $note . "
 <br>
 Item: " . $item . " <br>
 Item description: " . $itemdesc . " <br>
 Lunch item: " . $lunch . " <br>
 Dinner item: " . $dinner . " <br>
 Price: " . $price . " <br>
 </BODY>
 </HTML>";[/code]

@dvdd127authorNov 23.2007 — #check there is any error or not? by replace this line you know is there any error in the query or not

[code=php]
 
 mysql_query("INSERT INTO 'menu' ('category','item','itemdesc','price','lunch','dinner') VALUES ('$category','$item','$itemdesc','$price','$lunch','$dinner')") or die(mysql_error()); 
 [/code]

[/QUOTE]
i am not sure i understand what you are saying

Try a print_r($_POST) to see what values are set.[/QUOTE]
tried this... doesnt seem to fix the issue

Move the mysql_query() line down (after $dinner = $_POST['dinner']). The code should be as follows:
[/QUOTE]
tried this too and it also didnt work

I tried the last two suggestions both together and separate.... doesnt seem to work either way.

@roscorNov 24.2007 — #First confirm that your database connection info is correct,

best practice is to set those varibles eg dbinfo.php then use an include, like so include("dbinfo.php");

in that file you have your connection info

dbinfo.php

[code=php]$host = "localhost";
 
 $username = "chale";
 
 $password = "test";
 
 $database = "site";
 [/code]

[code=php]
 include("dbinfo.php");
 mysql_connect($host,$username,$password);
 @mysql_select_db($database) or die( "Unable to select database");
 $query = "INSERT INTO menu SET
 category = '$catergory',
 item = '$item',
 itemdesc = '$itemdesc',
 price = '$price',
 lunch = '$lunch',
 dinner = '$dinner',
 mysql_query($query);[/code]

maybee your intial problem was your admitance of a null value for your primary key which may be set as auto incremnent.

so another alternative could be

[code=php]$query =INSERT INTO landlordlo VALUES ('','$catergory','$item','$itemdesc','$price','$lunch','$dinner')";
 mysql_query($query);
 
 // then check if the info has been submited or display an error
 if($query){
 echo "<td align ='center'><font face='Verdana' size='2' color='#000000'>Record now added to Database<br><br><br>";
 
 die() ;
 }
 
 else {
 echo "<td align ='center'><font face='Verdana' size='2' color='#000000'>Record Failed to be added to database please try again";
 
 die() ;
 
 }[/code]

or something similar, hope this helps

@dvdd127authorNov 24.2007 — #First confirm that your database connection info is correct,

best practice is to set those varibles eg dbinfo.php then use an include, like so include("dbinfo.php");

in that file you have your connection info

dbinfo.php

[code=php]$host = "localhost";
 
 $username = "chale";
 
 $password = "test";
 
 $database = "site";
 [/code]

[code=php]
 include("dbinfo.php");
 mysql_connect($host,$username,$password);
 @mysql_select_db($database) or die( "Unable to select database");
 $query = "INSERT INTO menu SET
 category = '$catergory',
 item = '$item',
 itemdesc = '$itemdesc',
 price = '$price',
 lunch = '$lunch',
 dinner = '$dinner',
 mysql_query($query);[/code]

maybee your intial problem was your admitance of a null value for your primary key which may be set as auto incremnent.

so another alternative could be

[code=php]$query =INSERT INTO landlordlo VALUES ('','$catergory','$item','$itemdesc','$price','$lunch','$dinner')";
 mysql_query($query);
 
 // then check if the info has been submited or display an error
 if($query){
 echo "<td align ='center'><font face='Verdana' size='2' color='#000000'>Record now added to Database<br><br><br>";
 
 die() ;
 }
 
 else {
 echo "<td align ='center'><font face='Verdana' size='2' color='#000000'>Record Failed to be added to database please try again";
 
 die() ;
 
 }[/code]

or something similar, hope this helps[/QUOTE]
Thank you, but for some reason that did not work either.... any other ideas.

@dvdd127authorNov 24.2007 — #Ok. I manged to get it working. Now I just need to make it so that the message only comes up if the query is successful.

Here is the working code.

[CODE]<?
 
 mysql_connect("localhost", "binaryco_chale", "ale7");
 @mysql_select_db("binaryco_chale") or  die(mysql_error("Unable to connect to database."));
 
 $category = $_POST['category'];
 $item = $_POST['item'];
 $itemdesc = $_POST['itemdesc'];
 $price = $_POST['price'];
 $lunch = $_POST['lunch'];
 $dinner = $_POST['dinner'];
 
 $query = "INSERT INTO menu VALUES ('','$category','$item','$itemdesc','$price','$lunch','$dinner')";
 mysql_query($query);
 
 
 echo "The item has been added as stated below to the menu.<br>If there is a mistake you can edit it by using the search links at the top of the page for each section of the menu.";
 
 print ("<!-- " . $_POST['formaction'] . " -->
 <HTML>
 <HEAD>
 <TITLE>Adding menu item</TITLE>
 </HEAD>
 <BODY>" . $note . "
 <br>
 Item: " . $_POST['item'] . " <br>
 Item description: " . $_POST['itemdesc'] . " <br>
 Lunch item: " . $_POST['lunch'] . " <br>
 Dinner item: " . $_POST['dinner'] . " <br>
 Price: " . $_POST['price'] . " <br>
 </BODY>
 </HTML>");
 
 
 ?>[/CODE]

This is the message I would like to set to ONLY show up if the query is successful.

[CODE]echo "The item has been added as stated below to the menu.<br>If there is a mistake you can edit it by using the search links at the top of the page for each section of the menu.";
 
 print ("<!-- " . $_POST['formaction'] . " -->
 <HTML>
 <HEAD>
 <TITLE>Adding menu item</TITLE>
 </HEAD>
 <BODY>" . $note . "
 <br>
 Item: " . $_POST['item'] . " <br>
 Item description: " . $_POST['itemdesc'] . " <br>
 Lunch item: " . $_POST['lunch'] . " <br>
 Dinner item: " . $_POST['dinner'] . " <br>
 Price: " . $_POST['price'] . " <br>
 </BODY>
 </HTML>");
 
 
 ?>[/CODE]

Any ideas?? Thanks again for all your help.

@roscorNov 25.2007 — #use an if statement

[code=php]if(condition){
 something
 } else {
 something different
 }[/code]

therefore if your query has executed {display your message} else {display an error message}

see [URL]http://www.w3schools.com/php/php_if_else.asp[/URL]

@dvdd127authorNov 25.2007 — #use an if statement

[code=php]if(condition){
 something
 } else {
 something different
 }[/code]

therefore if your query has executed {display your message} else {display an error message}

see [URL]http://www.w3schools.com/php/php_if_else.asp[/URL][/QUOTE]thanks again for your help.

one final issue. a blank record is saved to the db whenever i load the page. I am sure this is once again something really easy. i am new to php though so that is why I dont know how to fix it.

@gpm1982Nov 26.2007 — #as suggested by roscor, you can use the if - else block to verify if the input has been submitted. example:

[code=php]
 if( isset($_POST['category']) ){
 mysql_connect("localhost", "binaryco_chale", "ale7");
 // the remaining code 
 }
 else{
 // display the form
 }
 [/code]

@dvdd127authorNov 26.2007 — #do i need to do that for each field.. or just the first field that is submitted?

@gpm1982Nov 26.2007 — #do i need to do that for each field.. or just the first field that is submitted?[/QUOTE]

You can check for every field submitted, or you can check for just one or two fields, it's up to you. For my case, I would normally check for one field. But then, I'll validate the other fields before I submit them to db. Example for user login page:

[code=php]
 if( isset($_POST['username']) ){
 // get the values, with every slashes stripped
 // this is not necessary, but I use mysql's built in escape character function
 $username = ( get_magic_quotes_gpc() ) ? stripslashes($_POST['username']) : $_POST['username'];
 $password = ( get_magic_quotes_gpc() ) ? stripslashes($_POST['password']) : $_POST['password'];
 
 // I want to make sure both username and password contains at min 6 to max 15 characters
 if( !preg_match("#^w{6,15}$#", $username) || !preg_match("#^w{6,15}$#", $password) ){
 // display error message, then exit
 }
 else{
 // continue with the script
 }
 }
 [/code]

@dvdd127authorNov 26.2007 — #thanks a lot for all your help.

Is there a way to make it so that on initial load it does not show an error message??

@gpm1982Nov 26.2007 — #thanks a lot for all your help.

Is there a way to make it so that on initial load it does not show an error message??[/QUOTE]

Can you tell me what error message appears when you perform initial load?

@dvdd127authorNov 26.2007 — #Can you tell me what error message appears when you perform initial load?[/QUOTE]sorry, i was not clear in what I was asking..

I was referring to my custom error message that lets the user know nothing was added to the db.

@dvdd127authorNov 27.2007 — #so is there a way to get rid of the custom error message that tells the user nothing was added to the db?

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