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up vote 1 down vote favorite

i have a simple problem with that fetch_array. i like to read out x1 and x2 from table A. the both arrays i need for a counter which will set up another field on another table B to +1. i wrote the following code but it doesnt work and i dont know where is the mistake. maybe someone can help me out. thanks alot.

$get_counter = "SELECT x1, x2 FROM tablename(A) WHERE id='$id'";
        $result2 = mysqli_query($db, $get_counter);

        $row = $result2->fetch_array(MYSQLI_ASSOC);

        $counter = "UPDATE tablename(B) SET xy=xy + 1 WHERE x1=$row["x1"] AND x2=$row["x2"]";
        $result3 = mysqli_query($db, $counter);
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What does not work? –  hakre Mar 12 '12 at 12:27
 
We assume tablename(A), tablename(B) are just placeholders here, as that is not a valid syntax. –  Michael Berkowski Mar 12 '12 at 12:31
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3 Answers

up vote 1 down vote accepted 

$counter = "UPDATE tablename(B) SET xy=xy + 1 WHERE x1='".$row["x1"]."' AND x2 ='".$row["x2"]."'";
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hello and thanks for helping me out. i had to do it that way to get it work. thanks to all of you. best wishes –  John Mar 12 '12 at 12:49
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up vote 2 down vote

Your quoting is faulty in $counter. It is good practice (and sometimes required) to surround arrays or objects in {} inside double-quoted strings.

$counter = "UPDATE tablename(B) SET xy = (xy + 1) WHERE x1={$row['x1']} AND x2={$row['x2']}";

However, if you do not intend to use x1, x2 aside from inside the second query, you can do this with one query and a JOIN. This eliminates the need for the first query and fetch call.

UPDTE 
 tablenameb B JOIN tablenamea A ON B.x1 = A.x1 AND B.x2 = A.x2
SET xy = (xy + 1)
WHERE A.id='$id'
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up vote 1 down vote

Your SQL contains an error and your script doesn't check for SQL errors.

This:

$row = $result2->fetch_array(MYSQLI_ASSOC);

Should be:

if ($result2 === false) {
    printf("Invalid query: %s\nWhole query: %s\n", mysqli_error(), $get_counter);
    exit();
}

$row = $result2->fetch_array(MYSQLI_ASSOC);

See how $result2 is checked here, and how the SQL error is printed if it failed.

share|improve this answer
 
i added that to all of the queries. finally it was displaying the mistake at result3. thanks alot. –  John Mar 12 '12 at 12:51
 
I'm glad it helped you, please close the question –  jpic Mar 12 '12 at 12:53
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