It's been driving me crazy...why is this returning a parse error? It says
the error is on the last line, I thought I was following the syntax....
<?php
$link = mysql_connect(" localhost");
if(! $link)
die("Could not connect to MySQL");
$database = "PC_Store";
mysql_select_db ($database)
or die ("could not open $database: ".mysql_err or() );
$Choice = $_POST['myradio'];
$db = mysql_connect(" localhost", "root");
mysql_select_db ("PC_Store",$db );
if ($Choice = 'Sales Data') {
?>
<BR>
Enter The Fields For Entry Into SalesData Table
<form action="<?echo $PHP_SELF?>" method=POST>
<input type="text" name="TransID"s ize=10>TransID
<BR><input type="text" name="ProductID "size=10>Produc tID
<BR><input type="text" name="TransCode "size=10>TransC ode
<BR><input type="text" name="Date"size =10>Date
<BR><input type="text" name="UnitsSold "size=10>UnitsS old
<input type="submit" name="submit" value="Submit Data">
</form>
<?php
$result = mysql_query("IN SERT INTO salesdata VALUES
($ProductID,$Tr ansCode,$Date,$ UnitsSold.)",$d b)
}
3  2866 
Got it working....but it won't save to the database...
<HTML>
<HEAD>
<TITLE>PC_Store </TITLE>
</HEAD>
<BODY BGCOLOR=RED>
<?php
$link = mysql_connect(" localhost");
if(! $link)
die("Could not connect to MySQL");
$database = "PC_Store";
mysql_select_db ($database)
or die ("could not open $database: ".mysql_err or() );
$Choice = $_POST['myradio'];
echo "<BR> you picked " . $_POST['myradio'] . "\n";
$db = mysql_connect(" localhost", "root");
mysql_select_db ("PC_Store",$db );
if ($Choice = 'Sales Data') {
?>
<BR>
Enter The Fields For Entry Into SalesData Table
<form action="<?echo $PHP_SELF?>" method=POST>
<input type="text" name="ProductID "size=10>Produc tID
<BR><input type="text" name="TransCode "size=10>TransC ode
<BR><input type="text" name="Date"size =10>Date (YYYY-MM-DD)
<BR><input type="text" name="UnitsSold "size=10>UnitsS old
<input type="submit" name="submit" value="Submit Data">
</form>
<?php
$result = mysql_query("IN SERT INTO salesdata VALUES
('$ProductID',' $TransCode','$D ate','$UnitsSol d')");
}
?>
<BR>
<FORM METHOD=POST ACTION = "PC_Store1.php" >
<INPUT TYPE = "submit" VALUE ="Return To Main Menu">
</FORM>
"Rob" a écrit le 24/11/2003 : It's been driving me crazy...why is this returning a parse error? It says
the error is on the last line, I thought I was following the syntax....
You forgot the final ; !
Last line becomes :
$result = mysql_query("IN SERT INTO salesdata VALUES
($ProductID,$Tr ansCode,$Date,$ UnitsSold.)",$d b);
> if ($Choice = 'Sales Data') {
More than likely, you probably want this to be
if ($Choice == 'Sales Data') {
(note the double equals). Your original code was setting $Choice to equal
'Sales Data', always returning true.
Chris Finke
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