Hello im making a browser game and have 99% working but iv got one error i cant find / fix and would like to know if anyone can help me where im going wrong heres the error - QUERY ERROR: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' 0, '1', '2121221', '1212121', '12121', '12121', 1211, '121', 121)' at line 1
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Query was INSERT INTO crimes (crimeNAME, crimeBRAVE, crimePERCFORM, crimeSUCCESSMUNY, crimeSUCCESSCRYS, crimeSUCCESSITEM, crimeGROUP, crimeITEXT, crimeSTEXT, crimeFTEXT, crimeJTEXT, crimeJAILTIME, crimeJREASON, crimeXP) VALUES( 'fdfdf', '1', '((WILL*0.8)/2.5)+(LEVEL/4)', '12', , 0, '1', '2121221', '1212121', '12121', '12121', 1211, '121', 121)
- '12', , 0, '1', '2121221', '1212121', '12121', '12121', 1211, '121', 121
this is where i would write but for the testing i did anything
here is the coding to the table in mysql - CREATE TABLE `crimes` (
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`crimeID` int(11) NOT NULL auto_increment,
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`crimeNAME` varchar(255) NOT NULL default '',
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`crimeBRAVE` int(11) NOT NULL default '0',
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`crimePERCFORM` text NOT NULL,
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`crimeSUCCESSMUNY` int(11) NOT NULL default '0',
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`crimeSUCCESSCRYS` int(11) NOT NULL default '0',
-
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`crimeSUCCESSITEM` int(11) NOT NULL default '0',
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`crimeGROUP` int(11) NOT NULL default '0',
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`crimeITEXT` int(11) NOT NULL default '0',
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`crimeSTEXT` int(11) NOT NULL default '0',
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`crimeFTEXT` int(11) NOT NULL default '0',
-
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`crimeJTEXT` int(11) NOT NULL default '0',
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`crimeJAILTIME` int(10) NOT NULL default '0',
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`crimeJREASON` varchar(255) NOT NULL default '0',
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`crimeXP` int(11) NOT NULL default '0',
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PRIMARY KEY (`crimeID`)
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) ENGINE=MyISAM ;
errors i get on page using - <?php
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error_reporting(E_ALL);
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ini_set('display_errors', true);
- Notice: Undefined index: HTTP_X_FORWARDED_FOR in /home/reddr431/public_html/smoke/header.php on line 111
-
-
Notice: Undefined index: attacking in /home/reddr431/public_html/smoke/header.php on line 120
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full code for this is attached below
been trying to fix this for a few weeks now
9  2308 
maybe the error is related to the type.Could be you are passing a integer or a string where MySQL is expecting the other type.
@guillermobytes
nope tryed changeing it around been looking alot of forums but still no answer
sorry i didn't look at the code, now i see the error, it's the VALUE of crimeSUCCESSCRY S that is not set in the query values.
It may be related to the missing index "attacking" on line 120 in /home/reddr431/public_html/smoke/header.php
the problem is not the table, its the query - Query was INSERT INTO crimes (crimeNAME, crimeBRAVE, crimePERCFORM, crimeSUCCESSMUNY, crimeSUCCESSCRYS, crimeSUCCESSITEM, crimeGROUP, crimeITEXT, crimeSTEXT, crimeFTEXT, crimeJTEXT, crimeJAILTIME, crimeJREASON, crimeXP) VALUES( 'fdfdf', '1', '((WILL*0.8)/2.5)+(LEVEL/4)', '12', MISSING_VALUE, 0, '1', '2121221', '1212121', '12121', '12121', 1211, '121', 121)
you have no value in crimeSUCCESSCRY S
see where i put MISSING_VALUE
ok done the input again with missing value and still have the error - QUERY ERROR:
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Query was INSERT INTO 'crimes' ('crimeNAME', 'crimeBRAVE', 'crimePERCFORM', 'crimeSUCCESSMUNY', 'crimeSUCCESSCRYS', 'crimeSUCCESSITEM', 'crimeGROUP', 'crimeITEXT', 'crimeSTEXT', 'crimeFTEXT', 'crimeJTEXT', 'crimeJAILTIME', 'crimeJREASON', 'crimeXP') VALUES( 'name', '1', '((WILL*0.8)/2.5)+(LEVEL/4)', '12', 0, 0, '1', 'intal text', 'success text', 'failure', 'jail', 12, 'reason', 12)
Dormilich 8,658
Recognized Expert Moderator Expert
maybe it’s because you want to pass a string to an integer field?
@Dormilich
problem fixed it was a simple code error
this
should have been
i was missing a '.' from the code all works now thanks for your help guys
@icemaster451
wow i have never seen that, plus the error output wasn't redirecting to this part of the code. wierd but glad you solved it
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