Multiple mysql_query not working

Question

I have a working mysql_query:

mysql_query("update products set buyers = buyers+$qtd where id=$pid")  or die (pgs_log("erro linha 70 >".mysql_error()));

But then I insert the following query right after it, and it only execute the first one:

mysql_query("update products set pending = pending-$qtd where id=$pid")  or die (pgs_log("erro linha 70 >".mysql_error()));

So, am I missing something?

try to echo sql queries to the screen and see whether $qtd and $pid is still working, if it is not, check your pgs_log (not sure what does it do). Is the query giving and error or is the page just loading the but database unchanged? — ahmet alp balkan
– ahmet alp balkan, Commented May 24, 2011 at 17:50
you have a SQL-injection hole in your code, change where id=$pid") into where id='$id'"). i.e. enclose all those $vars in single quotes or your mysql_real_escape_string's will be for naught. — Johan
– Johan, Commented May 28, 2011 at 21:05

onteria_ · Accepted Answer · 2011-05-24 17:51:24Z

1

Couple of things. First off you don't need two separate queries for this. MySQL may be confused into thinking your value is a column name because of the dash:

mysql_query("
UPDATE `products` 
SET `buyers` = `buyers` + $qtd, 
`pending` = `pending` - $qtd 
WHERE `id` = $pid")  or die (pgs_log("erro linha 70 >".mysql_error()));

answered May 24, 2011 at 17:51

onteria_

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Nick · Accepted Answer · 2011-05-24 17:54:28Z

1

f you wanted to double check that your update actually updated the data then you should investigate mysql_affected_rows. You'll need to check that your old value was different to your new value though, otherwise you'll have zero affected rows, making it a useless check.

You don't use proper quoting around the table and column references. These should be surrounded with back ticks, and could be combined, like the following:

UPDATE
    `products` 
SET
    `pending` = `pending` - $qtd,
    `buyers` = `buyers` + $qtd
WHERE
    `id` = $pid;

answered May 24, 2011 at 17:54

Nick

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Gaurav · Accepted Answer · 2011-05-24 17:56:29Z

1

mysql_query("update `products` 
             set `pending` = `pending` - $qtd, 
                 `buyers` = `buyers` + $qtd 
             where `id` = $pid")  
             or die (pgs_log("erro linha 70 >".mysql_error()));

edited May 24, 2011 at 17:56

answered May 24, 2011 at 17:50

Gaurav

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