What is wrong with this WHERE clause?

Is there a reason why this query doesn't work? The following query will work if I just exclude the WHERE clause. I need to know what is wrong with it. I know the given values of $key exists in the table, so why doesn't this work?

$q =   "SELECT * WHERE t1.project=$key
            FROM project_technologies AS t1
            JOIN languages AS t2
            ON t1.language = t2.key";

Table's have the following fields:

project_开发者_如何学Gotechnologies

- key

- project

- language

languages

- key

- name

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WHERE goes after FROM/JOIN.

SELECT    * 
FROM      project_technologies AS t1
JOIN      languages AS t2
ON        t1.language = t2.key
WHERE     t1.project=$key

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SELECT * FROM project_technologies AS t1
JOIN languages AS t2
  ON t1.language = t2.key
WHERE t1.project=$key

the where should be at the end (after JOINs)

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SELECT *
            FROM project_technologies AS t1
            JOIN languages AS t2
            ON t1.language = t2.key
            WHERE t1.project=$key

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In SQL you write :

SELECT ... FROM tables ... WHERE conditions

You put the stuff in the wrong order...

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The WHERE clause comes after the FROM clause.

SELECT *
FROM project_technologies as t1
JOIN languages as t2
on t1.language = ts.key
WHERE t1.project = $key

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I think WHERE has to come after FROM. Have you tried this?

$q =   "SELECT *
            FROM project_technologies AS t1
            JOIN languages AS t2
            ON t1.language = t2.key
            WHERE t1.project=$key";

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"SELECT * FROM project_technologies AS t1  INNER JOIN languages AS t2 ON t1.language = t2.key WHERE t.project = '$key'";

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FROM comes before WHERE in a SELECT statement.

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You should place your WHERE clause after your FROM clause.

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Maybe this should be the right way to write it

"SELECT * FROM project_technologies AS t1 JOIN languages AS t2 ON t1.language = t2.key WHERE t1.project=$key";