Updating variables

sjaakdelul · Nov 5, 2002

I am using php4.2.2 and the newest version of mysql.

Database:

CREATE TABLE cijferlijst (
Kwartaal int(1) default NULL,
vak varchar(5) default NULL,
cijfer1 varchar(4) default NULL,
cijfer2 varchar(4) default '-',
cijfer3 varchar(4) default '-',
cijfer4 varchar(4) default '-',
Studiepunten char(2) default '-',
Oordeel varchar(100) default 'Nog niet afgesloten'
)

I've one form:

<form name="form" method="post" action="update2.php">
<table width="500" border="0" cellpadding="5" cellspacing="0">
<?php
// MySQL data
$host = "localhost";
$username = "";
$password = "";
$database = "hrocijfers";

// Make connection to MySQL server
mysql_connect ("$host","$username", "$password"
mysql_select_db("$database"

// Execute query
$query = "SELECT * FROM `cijferlijst`";
$results = mysql_query($query);

// Create dropdown box
?>
<tr><td>Vak afkortingen:</td>
<td><div align="left"><? echo "<select name=vak>";
// List vakken into the box
while ($results_array = mysql_fetch_array($results)) {
$name = $results_array['vak'];
echo "<option value='" . $name . "'>" . $name . "</option>";
}
echo "</select>";
?></div></td>
</tr>
<tr><td>Afkorting vak:</td>
<td><div align="left">
<input type="text" name="vakkeuze" value="<?=$_POST['vakkeuze']?>">
</div></td></tr>
<tr><td>Behaalde cijfer:</td>
<td><div align="left">
<input type="text" name="cijferinvoer" value="<?=$_POST['cijferinvoer']?>">
</div></td></tr>
<tr><td>Hoeveelste cijfer:</td>
<td><div align="left">
<input type="text" name="cijfer" value="<?=$_POST['cijfer']?>">
</div></td></tr>
</table>
<input type="hidden" name="required" value="cijferinvoer">
<input type="submit" name="submit" value="submit">
</form>

Update2.php

<?php
if ( $_POST[vakkeuze]>A ) { $vakkeuze1=$_POST[vakkeuze];
echo "$vakkeuze1<br>";
} else {
echo "Fill in a good name!!!<br>";
}
// MySQL data
$host = "localhost";
$username = "upclan";
$password = "jXZDSuXw";
$database = "clanupclan";

// Make connection to MySQL server
mysql_connect ("$host","$username", "$password"
mysql_select_db("$database"

//Execute query
$query = "SELECT * FROM `cijferlijst` WHERE vak='$vakkeuze1'";
$results = mysql_query($query);

// Checkout the entered vak
$num_rows = mysql_num_rows($results);
if ( $num_rows>0 ) {
echo "password oke<br>";
}
else {
echo "Voer de juiste vak afkorting in.<br>";
}

if ( $_POST[cijferinvoer]>0 ) { $cijferinvoer1=$_POST[cijferinvoer];
echo "$cijferinvoer1<br>";
} else {
echo "voer een cijfer in dat groter is dan nul<br>";
}

if ( $_POST[cijfer]>0 ) { $cijfer1=$_POST[cijfer];
echo "$cijfer1<br>";
} else {
echo "voer een cijfer in dat groter is dan nul<br>";
}

// Execute query
$query = 'UPDATE cijferlijst set $cijfer1=$cijferinvoer where vak=$vakkeuze1';
$results = mysql_query($query);
echo "Ok";
?>

There are 5 steps.
1) Set $vakkeuze1
2) Set $cijferinvoer1
3) Set $cijfer1
4) Check vak
5) Update the table cijferlijst.

The steps 1,2 and 3 are going very well...but the steps 4 and 5 aren't good.

Step 4:
I've filled out the form:

Afkorting vak:Bekn1
Cijfer:8
Hoeveelste cijfer:1

The result from update2.php is:

Bekn1
Voer de juiste vak afkorting in.
8
1
Ok

So, the cause of the problem start at "// Checkout the entered vak" ...

But what's wrong?

Step5:
I think the code to update the table cijferlijst is good ... anyone who can see any wrong code?

Who can help me?

KarveR · Nov 5, 2002

use ' instaed of ` .
(`) is know to cause the odd problem with php, then try it again and see how you get on. ______________________________________________________________________
There's no present like the time, they say. - Henry's Cat.

sjaakdelul · Nov 5, 2002

Ok, thx, I'll try it tomorrow

sjaakdelul · Nov 5, 2002

Hmmz...disapointed, it didn't work

I create another file to test the update line. This works and I copy and past it into update2.php:

<?php
if ( $_POST[vakkeuze]>A ) { $vakkeuze1=$_POST[vakkeuze];
echo "$vakkeuze1<br>";
} else {
echo "Voer de juiste vakafkorting in!!!<br>";
}
// MySQL data
$host = "localhost";
$username = "";
$password = "";
$database = "hrocijfers";

// Make connection to MySQL server
mysql_connect ("$host","$username", "$password&quot

;
mysql_select_db("$database&quot

;

//Execute query
$sql = "SELECT * FROM 'cijferlijst' WHERE vak='$vakkeuze1'";
$result = mysql_query($sql);

// Checkout the entered vak
$num_rows = mysql_num_rows($result);
if ( $num_rows>0 ) {
echo "password oke<br>";
}
else {
echo "Voer de juiste vak afkorting in.<br>";
}

if ( $_POST[cijferinvoer]>0 ) { $cijferinvoer1=$_POST[cijferinvoer];
echo "$cijferinvoer1<br>";
} else {
echo "voer een cijfer in dat groter is dan nul<br>";
}

if ( $_POST[cijfer]>0 ) { $cijfer1=$_POST[cijfer];
echo "$cijfer1<br>";
} else {
echo "voer een cijfer in dat groter is dan nul<br>";
}
//Execute query
$query = 'UPDATE cijferlijst set cijfer1=9 where vak=\'Proj1\'';
$results = mysql_query($query);

echo "Your availability is updated... <br><br>";
echo "<a href=availability2.php>Click here to view the status of other members</a>";
?>

The update linew is correctly now

//Execute query
$query = 'UPDATE cijferlijst set cijfer1=9 where vak=\'Proj1\'';
$results = mysql_query($query);

-But how can I put the variables into the line?

$query = 'UPDATE cijferlijst set cijfer1='$cijferinvoer' where vak=\'Proj1\'';

this one didn't work...

-What can be wrong at the select line?

Thx

sjaakdelul · Nov 5, 2002

Ok, now I've this select line:

//Execute query
$sql = 'SELECT * FROM `cijferlijst` WHERE 1 AND `vak` = \'Proj1\' LIMIT 0, 30';
$result = mysql_query($sql);

Phpmyadmin advise me to use a variable in the line like this:

$sql = 'SELECT * FROM `cijferlijst` WHERE 1 AND `vak` = \'$vakkeuze1\' LIMIT 0, 30';

But the line is defect when I used it...

Anyone who knows more about this problem?

sleipnir214 · Nov 5, 2002

Try:

$sql = "SELECT * FROM cijferlijst WHERE 1 AND vak = ' . $vakkeuze1 . ' LIMIT 0, 30';

______________________________________________________________________
TANSTAAFL!

sjaakdelul · Nov 5, 2002

It didn't worked ... I've tried 10 other variants, but use a variable in the update or select line still didn't worked ...

Anyone knows more about this problem?

KarveR · Nov 7, 2002

if(!empty($vakkeuze)){

$sql = "SELECT * FROM cijferlijst WHERE vak = '$vakkeuze1' LIMIT 0, 30';

}else{
echo "vakkeuze1 is empty";
}

______________________________________________________________________
There's no present like the time, they say. - Henry's Cat.

sjaakdelul · Nov 7, 2002

This is the goodone:

//Execute query
$result = mysql_query("
SELECT * from cijferlijst
WHERE 1
AND vak = '". $vakkeuze1 . "'
&quot

;

Thx 4 all

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Updating variables

sjaakdelul

IS-IT--Management

KarveR

MIS

sjaakdelul

IS-IT--Management

sjaakdelul

IS-IT--Management

sjaakdelul

IS-IT--Management

sleipnir214

Programmer

sjaakdelul

IS-IT--Management

KarveR

MIS

sjaakdelul

IS-IT--Management

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