Parse error...

Ton_Wibier

Hi group!

As you can see, I'm a newbie. 🙁
Therefor, I have a question.

I want to pass variables from a HTML page to a PhP page. First let me show you the code in the HTML page:

<form action="queryplay.php" method="POST">
Zoekveld: <input type="text" name="zoekveld" />
Zoekterm: <input type="text" name="zoekterm" />
<br><br>
<input type="submit">
</form>

In my PhP form I want to make use of the red variables. Here I am using

<?php
mysql_connect("localhost","","");
mysql_select_db("vmeester");
$PostcodeLijst = mysql_query("SELECT * FROM $POST['zoekveld'] WHERE $POST['zoekterm']");
$Aantal = mysql_num_rows($PostcodeLijst);

"All is well!", you might think. Nope, this is not the fact. I'm getting the next error:

Parse error: parse error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in E:\queryplay.php on line 11

This is the bold line in the second quote.
What am I doing wrong here!? Please help!

planetsim

well whats line 11.. Thats the line the errors on

Ton_Wibier

Originally posted by planetsim
well whats line 11.. Thats the line the errors on

Good thinking! Now, what's the problem and how to solve it!?

planetsim

well can you show us lines 8 to 14..

Ton_Wibier

Okay, here it is:

<?php
mysql_connect("localhost","","");
mysql_select_db("vmeester");
$PostcodeLijst = mysql_query("SELECT * FROM $POST['zoekveld'] WHERE $POST['zoekterm']");
$Aantal = mysql_num_rows($PostcodeLijst);

if ($Aantal == 0):
$Aantal = "Geen records gevonden!";
elseif ($Aantal == 1):
$Aantal = $Aantal . " record gevonden.";
else:
$Aantal = $Aantal . " records gevonden.";
endif;
?>

There you have it!

BuzzLY

I suspect your problem lies in "zoekterm." Can you give us the values you typing into those two fields? Try taking the exact strings you are typing into the fields and hard coding the SQL query. Do you get the same error?

Your WHERE statement should say WHERE thisfield LIKE 'something' or WHERE thisfield = 'something.' If your zoekterm contains double quotes or no quotes at all, then it's probably causing your error.

Another thing to try when creating your SQL code is this:

$SQLString = "SELECT * FROM " . $POST['zoekveld'] . " WHERE " . $POST['zoekterm'];
$PostcodeLijst = mysql_query($SQLString);

This way, you can check the value of SQLString by echoing it to the screen, so you can check to make sure it's being created correctly.

Hope that helps.

Ton_Wibier

BuzzLY,

Thanks a lot for your answer. It works like a charm......for 95%! 🙁

For example, when I type postcode in the field "zoekveld" and PCOcode LIKE '%13%' in "zoekterm", I get this result:
SELECT * FROM postcode WHERE PCOcode LIKE \'%13%\'

This is almost right! The only thing is that MySQL does NOT accept the backslashes. And they are generated by PhP. Is there a way to prevent this?

dalecosp

Originally posted by Ton Wibier
This is almost right! The only thing is that MySQL does NOT accept the backslashes. And they are generated by PhP. Is there a way to prevent this?

Another newbie here....php has lots of functions for encoding. Take a look in the handbook at stripslashes(), htmlentities(),
htmlspecialchars(), etc., etc.,

As I'm SQL stupid I can't give you a specific answer....sorry!

HTH,

dalecosp

Ton_Wibier

Thanks for your suggestion. I've looked up the function and it runs like a charm! The slashes are stripped, indeed and gives me the result I want.

You made my day! 😃 😃

dalecosp

Originally posted by Ton Wibier
Thanks for your suggestion. I've looked up the function and it runs like a charm! The slashes are stripped, indeed and gives me the result I want.

You made my day! 😃 😃

Yippee!!!

I too am amazed how well things work when I use the right functions. I just wish I could memorize the dang manual....

dalecosp