Update Column Value From mysql_fetch_array result PHP

Question

I want to update the value of a column from a mysql table with the result from a mysql_fetch_array

But the last result is just the one being inserted/updated into the column

Where am i having wrong? Here's my code. Thanks in advance

    $studentname="some value";
    $course="some value";
    $query=mysql_query("select SABJEK,GRADE,REMARKS from table where STUDENTNAME='$studentname' && STUDENTNUMBER IS NULL") or die(mysql_error());
    while($result=mysql_fetch_array($query))
        {
        $sabjek=$result['SABJEK'];
        $grade=$result['GRADE'];
        $remarks=$result['REMARKS'];
        $msg="$sabjek = $grade - $remarks ";
        $msg1="$studentname $course $msg";
        }
    mysql_query("update table2 set `msg`='$msg1' where studentname='$studentname'") or die(mysql_error());

what are you expecting for $msg string?

Asik
– Asik

2014-11-02 08:29:44 +00:00
Commented Nov 2, 2014 at 8:29 — Asik
– Asik, Commented Nov 2, 2014 at 8:29
@Asik dont know man that's why i ask it here you have idea?

kristyan markes
– kristyan markes

2014-11-02 08:47:46 +00:00
Commented Nov 2, 2014 at 8:47 — kristyan markes
– kristyan markes, Commented Nov 2, 2014 at 8:47

motanelu · Accepted Answer · 2014-11-02 08:46:28Z

1

Put the mysql_query inside the while. Basically replace

  }
mysql_query("update table2 set `msg`='$msg1' where studentname='$studentname'") or die(mysql_error());

with

    mysql_query("update table2 set `msg`='$msg1' where studentname='$studentname'") or die(mysql_error());
}

and it will work :)

answered Nov 2, 2014 at 8:46

motanelu

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1 Comment

kristyan markes Over a year ago

did not work @motanelu the last result is just the one that was saved any idea? thanks

techy-coder · Accepted Answer · 2014-11-02 08:47:11Z

0

  If you want to update the column name every-time whenvere you fetch the result just put update query inside the while loop,   
   $studentname="some value";
   $course="some value";
   $query=mysql_query("select SABJEK,GRADE,REMARKS from table where STUDENTNAME='$studentname' && 
   STUDENTNUMBER IS NULL") or die(mysql_error());
   while($result=mysql_fetch_array($query));
     { $sabjek=$result['SABJEK'];
       $grade=$result['GRADE'];
       $remarks=$result['REMARKS'];
       $msg="$sabjek = $grade - $remarks ";
       $msg1="$studentname $course $msg";        
       mysql_query("update table2 set `msg`='$msg1' where studentname='$studentname'") or         
       die(mysql_error());
     }

if you are not expecting this result then please explain me in detail so, i will update the answer.

Thanks.

answered Nov 2, 2014 at 8:47

techy-coder

839 bronze badges

11 Comments

kristyan markes Over a year ago

there where 2 results from mysql_fetch_array result 1 and result 2 but result 2 is just the one inserted in the column why is it? i have placed update inside while thanks

kristyan markes Over a year ago

i want both result 1 and reult 2 to be inserted in the column thanks any idea @a.saleem

techy-coder Over a year ago

Fine, both results in the same column? if yes, could you please tell me what is the column data type?

kristyan markes Over a year ago

the column type of msg is text sir

techy-coder Over a year ago

Fine, you want to update the column please put the datatype as varchar,or let it be, consider it as example i am defining result1 is grade and result2 is course so , you want to show the values in the same column with separator eg with hyphen (-), $result1='A'; $result2='Maths'; So, Store it with $result3=$result1.'-'.$resultB; then if you want to show the result separetly just explode with (-) to show them as Grade & Course Thanks. Please dont call sir :) Cheers!

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