mysql_query() string?

[MertTheGreat]

hi,

I am a beginner site builder , I am reading Build Your Own Database Driven Website Using PHP & MySQL 3rd Edition by Kevin Yark.

I have an obstacle, and if you guys look at this code really quick and give me some answers, I safe time and would be really appreciated.

$jokelist = @mysql_query('SELECT joketext 
        FROM joke, author WHERE name=$name AND authorid=author.id'); 

this code does't read the variable "$name" how can I make it read it?
it gives this error:" Unknown column '$name' in 'where clause' "
it works if write the name manually.

solution:

$jokelist = @mysql_query("SELECT joketext 
        FROM joke, author WHERE name='$name' AND authorid=author.id"); 

Use double quotes not single quotes when you like to embed variables and have them evaluated in a string. If name is a string then you must also quote that.

Thanks

[lorenw]

I think you are missing something here, What are you expecting that to do?

It is just asking the database to get information where name=Tom (That is if $name='Tom';

With this in mind you need to set $name to someones name.

Then you have to run the query to get your results using a fetch, you will see where fetch is used in the book and will be run in a while loop.

[joaquin_win]

Use double quotes instead of single quotes:

$jokelist = @mysql_query("SELECT joketext
    FROM joke, author WHERE name=$name AND authorid=author.id"); 

[oddz]

$jokelist = @mysql_query("SELECT joketext 
        FROM joke, author WHERE name='$name' AND authorid=author.id"); 

Use double quotes not single quotes when you like to embed variables and have them evaluated in a string. If name is a string then you must also quote that.

[MertTheGreat]

thanks everyone it works sorry to bother

[cranial-bore]

At some point the error suppression (@) will cause you problems. If the query fails it will do so silently and you'll have a hard time figuring out what happened.

if(!$jokelist = mysql_query("SELECT ....")) {
     echo "Couldn't get jokes just now";
     //log mysql_error() for you to review
}