PHP Code Inserting Empty Records Into MySQL DB

Question

My code inserts an empty record into the MySQL table "table1" instead of getting what inserted in the field "Name" in form1.html. Any idea why it inserts an empty record instead of what the user entered in the field?

form1.html

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN"
    "http://www.w3.org/TR/html4/strict.dtd"
    >
<html lang="en">
<head>
    <title>Insert Your Name</title>
</head>
<body>
    <h3> Insert Your Name</h3>
    <form action="form1.php" method="post">
        <input type="text" name="Name">
            <input type="Submit" value="Submit" name="Submit">
    </form>
</body>
</html>

form.php

> <?php 
> $connection =
> mysql_connect("localhost","root","")
> or die ("Couldn't Connect To Server");
> $db = mysql_select_db("db1",
> $connection) or die ("Couldn't Select
> Database"); $query = "CREATE TABLE IF
> NOT EXISTS table1 (Name VARCHAR(20))";
> $result = mysql_query($query) or die
> ("Query Failed: " . mysql_error());
> $query = "INSERT INTO table1 (Name)
> VALUES ('$_post[Name]')"; $result =
> mysql_query($query) or die ("Query
> Failed: " . mysql_error()); $query =
> "SELECT * FROM table1"; $result =
> mysql_query($query) or die ("Query
> Failed: " . mysql_error());
>     echo "<TABLE BORDER = '1'>";
>     echo "<TR>";
>     echo "<TH>Name</TH>";
>     echo "</TR>";
>     
>     while ($row = mysql_fetch_array($result))
>     {
>         echo "<TR>";
>         echo "<TD>", $row['name'], "</TD>";
>         echo "</TR>";
>     }
>     echo "</TABLE>";
>     mysql_close($connection); ?>

+1 seldom newbie with reputation 1 able to post question with great readability :) — ajreal
– ajreal, Commented Dec 4, 2010 at 16:26
Thanks ajreal! I hope this is the place I'll be able to find answers:) — Tom Granot
– Tom Granot, Commented Dec 4, 2010 at 16:29
Have you looked in the database to verify if the name field is being saved there? That will determine if the problem is in your insert or your select. — Surreal Dreams
– Surreal Dreams, Commented Dec 4, 2010 at 16:36
Not sure I understand, Surreal Dreams. Do you mean that I hsould check if the Name field is saved properly in the database? because it is. — Tom Granot
– Tom Granot, Commented Dec 4, 2010 at 16:40

ajreal · Accepted Answer · 2010-12-04 16:33:18Z

2

Instead of

$query = "INSERT INTO table1 (Name) VALUES ('$_post[Name]')";

Try this

$query = "INSERT INTO table1 (Name) VALUES ('" .
          mysql_real_escape_string($_post['Name'],$db) . "')";

edited Dec 4, 2010 at 16:33

ajreal

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answered Dec 4, 2010 at 16:31

royas

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2 Comments

Tom Granot Over a year ago

Thanks to you to Roy-same question as I asked Pradeep: the php code that shows a table with the data from "table1" does not show the new recoed entered. Ideas?

royas Over a year ago

Propably it's the same problem: case.echo "<TD>", $row['Name'], "</TD>";

Horia Dragomir · Accepted Answer · 2010-12-04 16:36:21Z

0

 $query = "INSERT INTO table1 (Name)
 VALUES ('{$_POST[Name]}')"; $result =
 mysql_query($query) or die ("Query
 Failed: " . mysql_error());

$_POST and $_GET are uppercase.

change the display code to:

 while ($row = mysql_fetch_array($result))
     {
         echo "<TR>";
         echo "<TD>", $row['Name'], "</TD>";
         echo "</TR>";
     }

answered Dec 4, 2010 at 16:36

Horia Dragomir

2,87624 silver badges21 bronze badges

4 Comments

Tom Granot Over a year ago

Thanks to you to Horia! Wow, it's reall ynice how quickly stuff is answered around here. HoI will refer my second question to you too: the php code that shows a table with the data does not show the new recoed entered. Ideas?

Horia Dragomir Over a year ago

also, try to keep all things lowercase in the future. Less headaches that way.

Tom Granot Over a year ago

Works Perectly! Thanks Horia. So the problem is in the letter case, eh? Good to know.

Tom Granot Over a year ago

When I get the reputation, when I get the reputation:)

ajreal · Accepted Answer · 2010-12-04 16:32:21Z

0

Please change this

<input type="text" name="fname">


$query = "INSERT INTO table1 (Name)
 VALUES ('".mysql_real_escape_string($_POST[fname],$db)."')";

edited Dec 4, 2010 at 16:32

ajreal

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answered Dec 4, 2010 at 16:29

Pradeep Singh

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3 Comments

Tom Granot Over a year ago

However, there is a different problem now-the php code that shows a table with the data does not show the new recoed entered. Ideas?

Pradeep Singh Over a year ago

after insert please redirect the page

Tom Granot Over a year ago

Solved already, Pradeep, buy anyhow-what do you mean redirec the page? Should i build a new php file, and divide it to one that stores the data (ehich could be the HTML itself?) and one that shows the data?

Geeo · Accepted Answer · 2010-12-04 17:31:01Z

0

 " . ($_POST['Fname'] != "") ? $_POST['Fname'] : "No Name" ."

answered Dec 4, 2010 at 17:31

Geeo

112 bronze badges

1 Comment

Tom Granot Over a year ago

Thanks Geeo! What is the reason for the "No Name"? So if somebody inserted a blank field, that it would just write "no name"?

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