MySQL database update via form
Question
I'm making a form with which I can put something in my MySQL database, but for some reason it isn't working. When I try to submit the form it runs the else statement which means I get error Can someone tell me what I'm doning wrong???
I have two files the first one is the form:
<html>
<body>
<form name="form4" method="post" action="incident_submit.php">
<table>
<tr>
<td>Gebruiker</td>
<td>:</td>
<td>
<?php /* Include de pagina waar je config file staat */
include 'dbconnectie.php'; /* INCLUDE CONNECTION */
$query = $query = "SELECT naam FROM gebruikers"; /* YOUR QUERY */
$uitvoeren = mysqli_query($connectie,$query); /* EXECUTE QUERY */
echo "<select name='naam' id='naam'>";
while($rij = mysqli_fetch_array($uitvoeren)){ /* FETCH ARRAY */
/* De data in de dropdown zetten */
$naam=mysqli_real_escape_string($con,$rij['naam']);
echo "<option value='$naam'>".$naam."</option>";
}
/* De dropdown beëindigen */
echo '</select>';
?>
</td>
</tr>
<tr>
<td>Hardwaretypen</td>
<td>:</td>
<td>
<select name="typen" id="typen"> <?php
$query2 = "SELECT typen FROM hardware_typen"; /* YOUR SECOND QUERY */
/* Query uitvoeren */
$uitvoeren2 = mysqli_query($connectie,$query2); /* EXECUTE SECOND QUERY */
/* Begin van dropdown
While voor data in de dropdown */
while($rij2 = mysqli_fetch_array($uitvoeren2)) { /* SECOND FETCH */
$typen=mysqli_real_escape_string($con,$rij2['typen']); /* ESCAPE STRINGS */
/* De data in de dropdown zetten */
echo "<option value='$typen'>".$typen."</option>";
}
/* De dropdown beëindigen */
echo '</select>';
?>
</td>
</tr>
<tr>
<td>Melding</td>
<td>:</td>
<td><input name="melding" type="text" id="melding"></td>
</tr>
<tr>
<td><input type="submit" name="Submit" value="Incident melden"></td>
</tr>
</table>
</form>
</body>
</html>
The second file:
<?php
include('dbconnectie.php');
/* Get values from form */
$naam=mysqli_real_escape_string($con,$_POST['naam']); /* PRACTICE ESCAPING STRINGS BEFORE USING THEM INTO A QUERY */
$typen=mysqli_real_escape_string($con,$_POST['typen']);
$melding=mysqli_real_escape_string($con,$_POST['melding']);
/* Insert data into mysql */
$sql="INSERT INTO incidenten(naam, typen, melding) VALUES ('$naam', '$typen', '$melding')";
$result=mysqli_query($con,$sql); /* EXECUTE QUERY */
// if successfully insert data into database, displays message "Successful".
if($result){
echo "Uw melding is verwerkt in het systeem.";
echo "<BR>";
echo "<a href='gebruikerpagina.html'>Terug naar de startpagina</a>";
}
else {
echo "ERROR";
}
?>
<?php
// close connection
mysqli_close($connectie);
?>
My MySQL table gebruikers:
CREATE TABLE IF NOT EXISTS `gebruikers` (
`naam` text NOT NULL,
`wachtwoord` text NOT NULL,
`level` int(1) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
INSERT INTO `gebruikers` (`naam`, `wachtwoord`, `level`) VALUES
('niek', 'P@password1', 1),
('Hansie', 'Hansie1', 3);
Thanks in advance!
4 answers
solution1
0 2014-06-02 07:21:56
Try
$sql="INSERT INTO {$tbl_name}(naam, typen, melding)VALUES('$naam', '$typen', '$melding')";
$result=mysql_query($sql) or die(mysql_error()); // will display the error
When adding - or die (mysql_error()) it will display errors related to MySql engine (such as wrong username & password, typos , etc)
solution2
0 2014-06-02 07:22:53
add name and Id to select for both naam and typen
echo '<select id="naam" name="naam">';//add name and id
//While voor data in de dropdown
while($rij = mysql_fetch_assoc($uitvoeren)) {
//De data in de dropdown zetten
echo '<option>'.$rij['naam'].'</option>';
}
//De dropdown beëindigen
echo '</select>';
and here
echo '<select id="typen" name="typen">';//add name and id
//While voor data in de dropdown
while($rij = mysql_fetch_assoc($uitvoeren)) {
//De data in de dropdown zetten
echo '<option>'.$rij['typen'].'</option>';
}
//De dropdown beëindigen
echo '</select>';?>
solution3
0 2014-06-02 07:23:07
change your form select to ( you have one more select in select ) Also no need to include file twice:-
<td>
<?php //Include de pagina waar je config file staat
include 'dbconnectie.php';
//Query
$query = "SELECT naam FROM gebruikers WHERE level='3'";
//Query uitvoeren
$uitvoeren = mysql_query($query);
?>
<select name="naam" id="naam" >
<?php while($rij = mysql_fetch_assoc($uitvoeren)) {?>
<option value="<?php echo $rij['naam'];?>"><?php echo $rij['naam'];?></option>
<?php }?>
</select></td>
</tr>
<tr>
<tr>
<td>Hardwaretypen</td>
<td>:</td>
<td>
<?php $query = "SELECT typen FROM hardware_typen";
//Query uitvoeren
$uitvoeren = mysql_query($query);?>
<select name="typen" id="typen">
<?php while($rij = mysql_fetch_assoc($uitvoeren)) {?>
<option value="<?php echo $rij['typen'];?>"><?php echo $rij['typen'];?></option>
<?php }?>
</select></td>
and then insert query to :-
$sql="INSERT INTO incidenten (naam, typen, melding)VALUES('$naam', '$typen', '$melding')";
solution4
0 ACCPTED 2014-06-02 07:37:34
Summary of Answer:
- Converted code from deprecated MySQL to MySQLi .
- Too many include function of dbconnectie.php where-in you can just use it once in a page.
- Tried to clean the html table.
- Used mysqli_real_escape_string function on variables to prevent some of the SQL injections.
- Removed unnecessary and excessive
<select> - Missing
> - And importantly ! Missing value tags in your
<option>
I have included explanations quoted in /* */ inside the code I have provided for further reference.
dbconnectie.php :
<?php
/* ESTABLISH CONNECTION */
$con=mysqli_connect("localhost","root","","helpdesk_middenpolder");
if(mysqli_connect_errno()){
echo "Error".mysqli_connect_error();
}
?>
Your form :
<html>
<body>
<form name="form4" method="post" action="incident_submit.php">
<table>
<tr>
<td>Gebruiker</td>
<td>:</td>
<td>
<?php /* Include de pagina waar je config file staat */
include 'dbconnectie.php'; /* INCLUDE CONNECTION */
$uitvoeren = mysqli_query($con,"SELECT * FROM gebruikers"); /* YOUR QUERY */
echo "<select name='naam' id='naam'>";
while($rij = mysqli_fetch_array($uitvoeren)){ /* FETCH ARRAY */
/* De data in de dropdown zetten */
$naam=mysqli_real_escape_string($con,$rij['naam']);
echo "<option value='$naam'>".$naam."</option>";
}
/* De dropdown beëindigen */
echo '</select>';
?>
</td>
</tr>
<tr>
<td>Hardwaretypen</td>
<td>:</td>
<td>
<select name="typen" id="typen">
<?php
$query2 = "SELECT typen FROM hardware_typen"; /* YOUR SECOND QUERY */
/* Query uitvoeren */
$uitvoeren2 = mysqli_query($con,$query2); /* EXECUTE SECOND QUERY */
/* Begin van dropdown
While voor data in de dropdown */
while($rij2 = mysqli_fetch_array($uitvoeren2)) { /* SECOND FETCH */
$typen=mysqli_real_escape_string($con,$rij2['typen']); /* ESCAPE STRINGS */
/* De data in de dropdown zetten */
echo "<option value='$typen'>".$typen."</option>";
}
/* De dropdown beëindigen */
echo '</select>';
?>
</td>
</tr>
<tr>
<td>Melding</td>
<td>:</td>
<td><input name="melding" type="text" id="melding"></td>
</tr>
<tr>
<td><input type="submit" name="Submit" value="Incident melden"></td>
</tr>
</table>
</form>
</body>
</html>
incident_submit.php :
<?php
include('dbconnectie.php');
/* Get values from form */
$naam=mysqli_real_escape_string($con,$_POST['naam']); /* PRACTICE ESCAPING STRINGS BEFORE USING THEM INTO A QUERY */
$typen=mysqli_real_escape_string($con,$_POST['typen']);
$melding=mysqli_real_escape_string($con,$_POST['melding']);
/* Insert data into mysql */
$sql="INSERT INTO incidenten (naam, typen, melding) VALUES ('$naam', '$typen', '$melding')";
$result=mysqli_query($con,$sql); /* EXECUTE QUERY */
// if successfully insert data into database, displays message "Successful".
if($result){
echo "Uw melding is verwerkt in het systeem.";
echo "<BR>";
echo "<a href='gebruikerpagina.html'>Terug naar de startpagina</a>";
}
else {
echo "ERROR";
}
// close connection
mysqli_close($con);
?>
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