Hello,
Notice: Undefined variable: ek_yuk in
$ek_yuk += $table['DATA_FREE'];
How do I fix this error?
By making sure the variable exists before you try and use it.
have you tried php isset() function to check.
or coalesce operator.
I try this way but it gives the same error
$result = $mysqli->query("
SELECT TABLE_SCHEMA,
TABLE_NAME,
(INDEX_LENGTH+DATA_LENGTH) AS SIZE_KB,
TABLE_ROWS, DATA_FREE
FROM INFORMATION_SCHEMA.TABLES
WHERE TABLE_SCHEMA = '$db_name'
ORDER BY $sort");
while($table = $result->fetch_array()) {
if(isset($table['DATA_FREE'])){
@$ek_yuk += $table['DATA_FREE'];
$ekyuk = showSize($ek_yuk);
}
$boyutu = showSize($table['SIZE_KB']);
@$satir += $table['TABLE_ROWS'];
$satirlar = number_format($satir, 0, ',', '.');
@$boyut += $table['SIZE_KB'];
$toplam = showSize($boyut);
}initialize that variable before making operations on it.
I’m a beginner, You can explain a little more
You set a value to the variable before using it the first time.
What is the real problem you are trying to solve with this? It seems rather odd that a beginner is going to query the information schema.
this is the same as
$ek_yuk = $ek_yuk + $table['DATA_FREE'];
The left $ek_yuk will be assigned with a new calculated value which is fine.
The right $ek_yuk (after the =) should be an existing variable while it is not (on the first time).
Wrong:
<?php
while($i < 10) { // you want to compare a non existing variable
$i++;
}
?>
Right:
<?php
$i = 0; // add a new variabele $i with a initial value of 0.
while($i < 10) { // you compare an existing variable now
$i++;
}
?>Thank you so much
The “Notice: Undefined variable” issue has improved as follows
$ek_yuk = 0;
$satir = 0;
$boyut = 0;
while($table = $result->fetch_array()) {
$ek_yuk = $ek_yuk += $table['DATA_FREE'];
$ekyuk = showSize($ek_yuk);
$boyutu = showSize($table['SIZE_KB']);
$satir = $satir += $table['TABLE_ROWS'];
$satirlar = number_format($satir, 0, ',', '.');
$boyut = $boyut += $table['SIZE_KB'];
$toplam = showSize($boyut);According to option “error2” gives the “Notice: Undefined variable: error2” error
if (!$error && !$error2 && $optimize)
How can I fix this error?
NOTE: I received this code from the internet immemorial, I can not write such a code,
// Loop Tables
foreach ($tables AS $table){
$error = 0;
$optimize = 1;
// Check Table
$check = $mysqli->query("CHECK TABLE `$table`");
while ($status = $check->fetch_array()){
// Status
if ($status[2] == 'error')
{
if ($status[3] == 'The handler for the table doesn\'t support check/repair')
{
$optimize = 0;
}
else
{
$error = 1;
}
}
}
// Check Table Error
if ($error)
{
// Repair Table
$repair = $mysqli->query("REPAIR TABLE `$table`");
// Status
if ($repair[3] != 'OK')
{
$error2 = 1;
}
else
{
$error2 = 0;
$error = 0;
}
}
// Check Optimize
if (!$error && !$error2 && $optimize)
{
// Optimize Table
$optimize = $mysqli->query("OPTIMIZE TABLE `$table`");
while ($status = $optimize->fetch_array())
{
// Status
if ($status[2] == 'error')
{
$error = 1;
}
}
}
}define that variable before using it?
I could not understand, Can you give sample?
$error = ‘Some strange error’; // define or initializing the variable
echo $error; // use the variable
Above goes well Adem but if you forget the first line and just do:
echo $error; // use the variable
What should the variable contain for content? you will get a Notice: Undefined variable: error
Thank you, problem is improved
foreach ($tables AS $table){
$error = 0;
$error2 = 1;
$optimize = 1;
// Check Table