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#1
August 4th, 2004, 04:11 PM
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Parse error
Hey, Im a total newbie, and I was wondering if anyone could suggest why this line of php is giving me a parse error, and suggest possible solutions.
 if((substr_count($Urow['user_items'],"ß".$row2['itemneeded']."fi") > 0) && ($ITrow['battle_effect'] <= $Urow['user_statlevel']) && ($ITrow['req_att'] <= $Urow['user_att']) && ($ITrow['req_def'] <= $Urow['user_def']) && ($ITrow['req_mst'] <= $Urow['user_mst']) && ($ITrow['req_ata'] <= $Urow['user_ata']) && ($ITrow['req_evp'] <= $Urow['user_evp']) && ($ITrow['class'] LIKE '%"."ß".$Urow[user_class]."fi"."%' OR $ITrow['class'] = 'ßALLfi'))
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#2
August 5th, 2004, 02:45 AM
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Yes its formatted horribly
 I suggest spacing it out some, put the different if conditions on different lines, and place what is run if the condition is true on a line by itself. That will go a long way to helping you find the problem. -KM-
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#3
August 5th, 2004, 05:13 AM
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What kode_monkey means is making it look like this:
PHP Code:
Now we easily see the problem, which is on the last row of the statement. PHP does not know the 'LIKE' statement, which comes from SQL. I assume that some code is missing between the $ITrow['class'] and the LIKE, a quote (") if nothing else. Good luck
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