how can i get mysql record like a data?

Question

Hi to all my question is some weird but i'll explane in here.

i want to record cv examples and user informations. i have two table like that;

cv
*id *format
users
*id *name *age

in format cell include some html tags and $name , $age tags like that;

<p>
My name is <b>$name</b> and im <b>$age</b> years old.
<p>

my codes are like that;

$sqlcv = mysql_fetch_array(mysql_query("select * from cv where id='2'"));
$text = $sqlcv['format'];

$sqluser = mysql_fetch_array(mysql_query("select * from users where id='3'"));
$name = $sqluser['name'];
$age = $sqluser['age'];

echo $text;

result; My name is $name and im $age years old. so html tags works but name and age not shown :( hope i can explane this one

That works exactly as is. Are you sure you're not using single quotes? — Kisaragi, Jul 12 '16 at 13:25
Possible duplicate of [php - insert a variable in an echo string](http://stackoverflow.com/questions/8054989/php-insert-a-variable-in-an-echo-string) — Himanshu Tyagi, Jul 12 '16 at 13:25
If any of the answers helped you, don't forget to mark it as Accepted. — Kalko, Sep 02 '16 at 10:55

score 2 · Answer 1 · answered Jul 12 '16 at 13:26

2

You can use sprintf()

echo sprintf("Hi im %s and im %d years old.", "Darius", $age);

answered Jul 12 '16 at 13:26

Indrasis Datta

8,692
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score 0 · Answer 2 · answered Jul 12 '16 at 13:24

0

Try this:

$age = 23;
$text = "Hi im Darius and im ".$age." years old";
echo $text;

answered Jul 12 '16 at 13:24

Dhara Parmar

8,021
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score 0 · Answer 3 · answered Jul 12 '16 at 13:25

0

You have to write $text like this:

$text = "Hi im Darius and im " . $age . " years old";

Else it will think $age is in the string.

answered Jul 12 '16 at 13:25

kerv

315
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score 0 · Answer 4 · edited Aug 31 '16 at 16:18

<p>My name is <b>$name</b> and I'm <b>$age</b> years old.<p>

I can't tell from your question, but you might have forgotten to wrap it in php. If that's the case, try using

<?php echo "<p>My name is <b>$name</b> and I'm <b>$age</b> years old.<p>"; ?>

or

<p>My name is <b><?=$name?></b> and I'm <b><?=$age?></b> years old.<p>

OR if you actually have the string "My name is $name and I'm $age years old." in the database cell (I hope you don't) and want to replace the variables with custom values, I suggest using

$output = str_replace('$age', $yourAgeVariable, $theCellValue)

and similarly with $name. I think that wouldn't make sense to put that into a database column though, so I'm posting this other part just in case you actually did that.

how can i get mysql record like a data?

4 Answers4