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up vote 2 down vote favorite

I am trying to get a variable to display, when used as part of a value inside of a mysql table.

$variable = 'cool';

MYSQL FIELD VALUE

"This is '.$variable.' wow"

When i echo this value, it displays as:

This is '.$variable.' wow

I want it to display as:

This is cool wow

What is the trick here?

@Mark sorry im new to this

$linkQuery3 = 'SELECT model
FROM models
WHERE model_id = "'.$pageModel.'"
';
$sql15 = mysql_query($linkQuery3) or die(mysql_error());
if (mysql_num_rows($sql15) == 0) {
    die('No results.');
} else {
    while($row = mysql_fetch_assoc($sql15)) {
     $model = stripslashes($row['model']);
    }
}

$linkQuery2 = 'SELECT l.link , l.desc , l.domId , d.domain FROM links l INNER JOIN domains d ON d.domId = l.domId WHERE l.catId="'.$pageCat.'" && (l.modId="1" || l.modId="'.$pageModel.'") ORDER BY domain
';
$sql3 = mysql_query($linkQuery2) or die(mysql_error());
if (mysql_num_rows($sql3) == 0) {
    die('No results.');
} else {
    $pageContent .= '';
    while($row = mysql_fetch_assoc($sql3)) {
     $linkAd = ($row['link']);
     $linkDesc = ($row['desc']);
     $linkDomain = ($row['domain']);
     $pageContent .= '
        <li><a href="'.$linkAd.'" target="_tab">'.$linkDesc.' '.$linkDomain.'</a></li>
    ';
    }
}
share|improve this question
1  
Looks like your quote marks are mismatched. Try setting the db value to "This is ".$variable." wow" –  Matt H. Jul 12 '11 at 17:18
    
tried this and displays as ".$variable." quotes are like that because echo'd field is displayed within another variable, by chance has it got something to do with some kind of declaration? like htmlspecialchars –  IndigoIdentity Jul 12 '11 at 17:24
    
@Mark alright thanks for the edit, that was driving me crazy. What i am trying to do is get the value of $model to show up in the value of the desc field when $linkDesc is echo'd –  IndigoIdentity Jul 12 '11 at 18:19

5 Answers 5

up vote 2 down vote accepted

What you are doing here is trying to echo the string from the database, but replace placeholder text with a specific value. You cannot do this by just storing a string that the PHP parser would treat in a specific way, and expect PHP to treat it the same way when it sees that string at run-time.

Here is what I suggest: Use a more straightforward delimiter for the part of the string you wish to replace, like so:

"This is :variable: wow"

and use str_replace() to echo the right thing. (You can also use sprintf() for the same purpose.)

echo str_replace(':variable:', $variable, $mystring);

(Here $mystring contains the string from the database.)

share|improve this answer
    
sounds like we're both on the same page here, let me let this info soak in and i'll let you know :) thank you! –  IndigoIdentity Jul 12 '11 at 17:31
    
how would str_replace work in this case? $linkAd = ($row['link']); the variable is inside of a value in the field link –  IndigoIdentity Jul 12 '11 at 17:49
    
I'm not sure what you're asking. Are you saying that you want to retrieve the column link and replace a placeholder that appears within? Just use $linkAd = str_replace(':variable:', $variable, $row['link']). Or are you saying that you want to use the value of link to perform the substitution? Use echo str_replace(':variable:', $row['link'], $mystring). –  Hammerite Jul 13 '11 at 17:02
    
Had to use preg_replace –  IndigoIdentity Aug 5 '11 at 21:24
up vote 2 down vote

use double quotes :

echo "This is " . $variable . " wow";
share|improve this answer
    
displays as " . $variable . " –  IndigoIdentity Jul 12 '11 at 17:22
1  
OP has the string as literal text in his database. echoing a variable containing the string does not magically make it be treated as PHP code. –  Marc B Jul 12 '11 at 17:29
up vote 2 down vote

You'd need to eval() the string you retrieve from MySQL. There's no other way of getting a string to be interpreted as PHP code. eval() should be avoided at all costs, however. You'd be better off using a templating system, or a simple string-substitution system, than eval()ing stuff coming out of a database.

share|improve this answer
    
This is correct. +1. Another question would be, why are you storing the variable in the database? Shouldn't you store the variable contents in the database and reverse the way you're storing/displaying things? –  rockerest Jul 12 '11 at 17:25
    
I don't see a reason to suggest eval() here, when nothing in the OP suggests he wants to do anything other than echo parameterised strings. –  Hammerite Jul 12 '11 at 17:28
    
eval() expects valid JS code. You'd need to do $var = 'cool'; $string = 'This is $var'; eval("\$newstring = '$string'");. But again, eval() is evil. You're not doing simple string substitution, you're trying to write a whole new PHP script on the fly. It's a bad idea. use str_replace() and the like instead. –  Marc B Jul 12 '11 at 17:31
1  
@Hammerite: yes, but OP is doing INSERT INTO table (x) VALUES ('This is $variable'), so that the literal word $variable is in the database. There's no way to get the retrieved string to be interpreted and $varaible substituted with its value without eval, unless you want to do str_replace($string, '$variable', 'cool'); –  Marc B Jul 12 '11 at 17:33
    
See the variable contents are in the database "$model = stripslashes($row['model']);" This is why its a little more complicated than my example. I was thinking stripslashes needs to be replaced, as said, by a directive saying that the field link contains php code being $model? –  IndigoIdentity Jul 12 '11 at 17:35
up vote 1 down vote

Try this:

"This is '" . $variable . '" wow"
share|improve this answer
    
displays as '" . $variable . '" –  IndigoIdentity Jul 12 '11 at 17:21
up vote 0 down vote

This can be a heck of a lot simpler. Just do this...

echo "This is '$variable' wow";

That is all you need. There is no need to leave the string context and come back in. The outer double-quotes will evaluate the value of a variable. If you had used single quotes on the outside then it would be evaluated literally as $variable and not its value.

So you would write a select statement like this:

$sql = "SELECT * FROM fancytbl WHERE id = '$id' AND myhands = 'diamonds'";

See, its simple.

share|improve this answer
    
its more complicated than my little example i promise :) the basic idea is there though. –  IndigoIdentity Jul 12 '11 at 17:31

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