Mysql value as parameter

[photocroatia]

Hi All,

I am very new to php and mysql.

I have the following query which works fine

$settings = mysql_query("SELECT * FROM table_name WHERE column_id = $var_no") or die("query failed");

$settings = mysql_fetch_array($settings);

echo $settings["description"]

-----------

The problem I have is I want to use that value that is returned eg. echo $settings["description"] as a parameter when calling another php script

@include ("http://www.mysite.com/body.php?myplace=NEW_VARIABLE_RETURNED");

NEW_VARIABLE_RETURNED is where the parameter should go I tried a lot of options without success.

Thank you

[dreamcatcher]

Hi,

Firstly, if you are only pulling one row of data, mysql_fetch_row or fetch_object are better. Always put your query between apostrophes. Also, best not to use a variable name as one you are already using. This can cause problems.

Try:

$settings = mysql_query("SELECT * FROM table_name WHERE column_id = '$var_no' LIMIT 1") or die("query failed");
$row = mysql_fetch_object($settings);

@include ("http://www.mysite.com/body.php?myplace=" . $row->description);

This should work ok. Just remember to use the include after you have pulled the data. I have also added the LIMIT clause, which is a good idea if you are only pulling one row.

:)

[baertyp]

Photocroatia,

by include()ing a file the parser replaces the include() statement with the contents of the referenced file.

An example:

<?php 
 // script1.php 
 $var1 = "foo"; 
 $var2 = "bar"; 
 include("script2.php"); 
?> 

<?php 
 // script2.php 
 echo $var1 . $var2; 
?> 

internally becomes

<?php 
 $var1 = "foo"; 
 $var2 = "bar"; 
 echo $var1 . $var2 
?> 

and will of course write

foobar

as naturally all of the variables defined in script1.php are also valid for script2.php.

What you are looking for most probably is:

// query goes here, result in $result 
header("Location: [some.whe.re...]  

which points the browser to the given URL.

Regards
Markus

[edited by: baertyp at 1:56 pm (utc) on Nov. 13, 2004]

[photocroatia]

I changed the code as per your instructions and did echo $row to see what it is returing,it echoed "Object" , this value does not exist in the table, am I still confused?

Thank you.

[dreamcatcher]

Are you using the object operator?

$row = mysql_fetch_object($settings);
echo $row->description;

Should work fine if the description row exists. Alternatively, you can try mysql_fetch_row.

$row = mysql_fetch_row($settings);

echo $row[0];

This assumes that the description row is your first row. If not, change the numerical indice.

dc :)

[photocroatia]

Baertyp,

The variable bit I understand and passing that to the calling php script.

In my original example when I echo the value eg.

echo $settings["description"]

This displays the value I want but I cant pass that to the calling php eg.

@include ("http://www.mysite.com/body.php?myplace=$settings["description"]");

I also can’t seem to assign that to a variable eg.

$newval=$settings["description"]

and then pass that to the calling script:

@include ("http://www.mysite.com/body.php?myplace=$newval");

Hopefully it is a little clearer now, sorry its hard to explain.

[photocroatia]

Dreamcatcher

Thank you that worked!

$row = mysql_fetch_row($settings);
echo $row[0];

I then used:

@include ("http://www.mysite.com/body.php?myplace=$row[0]”);

Thank you very much.

[dreamcatcher]

welcome. :)

[bobnew32]

Hmmm just a question about the piece of code, what does putting the "@" in front of a line of code do? I see it sometimes at different connections and stuff. Thx.

[jatar_k]

Error Control Operators [php.net]

PHP supports one error control operator: the at sign (@). When prepended to an expression in PHP, any error messages that might be generated by that expression will be ignored.

[photocroatia]

I have hit an issue now, if somebody could help please.

The issue is when $row[0] is 2 words, if I echo $row[0] both words are displayed, but when I call my other script:

@include ("http://www.mysite.com/body.php?myplace=$row[0]”);

It doesn't work, doesn't run, I have tried single quoates in various places but no luck.

Any assistance is appreciated.

Thank you

[dreamcatcher]

Maybe try urlencode() [uk2.php.net].

@include ("http://www.mysite.com/body.php?myplace=urlencode($row[0])");

[photocroatia]

Thank you, I tried urlencode() but it didn't work, I camr up with a work around, I moved my sql to the second script and just pased the ID to it, that worked fine.

Thanks again!

[Slade]

This is for any of the guru's out there following this thread.

Doesn't include'ing from an http add an extra level of indirection and processing to the execution of a page?

It appears the OP isn't trying to do a redirect, or I would have expected them to latch onto the header(...) example that was posted.

How about just wrapping the code in the include file into a function, and calling it like:

include('body.php');
doBody($row[0]);

At least then, if the code was needed more than once, you wouldn't have to call the page again, you'd just doBody() with the next variable.