PHP issues the error undefined offset error

Question

I want to compare two MySQL tables and to drop columns of table 2 if not in table 1. The code I am using is

<?php 

mysql_connect("localhost", "root", "") or die(mysql_error()); 
mysql_select_db("test") or die(mysql_error()); 


$fields = array();
$fields2 = array();
$dropcols = array();

$res=mysql_query("SHOW COLUMNS FROM table1");
$res2=mysql_query("SHOW COLUMNS FROM table2");

while ($x = mysql_fetch_assoc($res)) {
    $fields[] = $x['Field'];
}

while ($x = mysql_fetch_assoc($res2)) {
    $fields2[] = $x['Field'];
}

$diff = array_diff($fields2,$fields);

$arraylen = count($diff);

for ($x=0; $x < $arraylen; $x++) {
    mysql_query("ALTER TABLE table2   DROP  $diff[$x]");
}

Some times the code working and sometimes issue the error undefined offset 0. I don't know where the error is.

Thank you Barmar, Worked like a charm. I am a beginner in programming. — Bala
– Bala, Commented Nov 16, 2013 at 12:13
Excuseme Barmar for bothering u. Could u explain the usage of foreach(..) , $diff = array_values(..). Or Direct me to a link to study these in details if possible. Thank u. — Bala
– Bala, Commented Nov 16, 2013 at 12:24

Barmar · Accepted Answer · 2013-11-16 10:57:55Z

2

array_diff doesn't reindex the elements of the resulting array, they keep their indexes from fields, so there are gaps in the index. Use foreach to loop through the array values regardless of the indexes.

foreach ($diff as $column) {
    mysql_query(  "ALTER TABLE table2   DROP  $column");
}

Or you can do:

$diff = array_values(array_diff($fields2, $fields));

answered Nov 16, 2013 at 10:57

Barmar

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Comments

Nisanth Sojan · Accepted Answer · 2013-11-16 10:57:49Z

-1

it is becouse sometimes the array $diff will not contain any values

answered Nov 16, 2013 at 10:57

Nisanth Sojan

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1 Comment

Barmar Over a year ago

In that case $arraylen will be zero and the loop won't do anything.

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