I am getting Notice: Undefined variable: i in C:\xampp\htdocs\search-engine\search\search.php on line 23

Question

     foreach ($terms as $each) {

         $i++;

         if ($i == 1)
             $query .= "keywords LIKE '%$each%' ";  
         else
             $query .= " OR keywords LIKE '%$each%' ";

     }

you did not defined the i variable. if defined then please show full code. — Amir Iqbal, Oct 09 '15 at 04:08
When it is clearly said that is is not defined then did you tried looking at where you have defined $i before doing increment in its value — Arpita, Oct 09 '15 at 05:42

Qaisar Satti · Answer 1 · 2015-10-09T04:53:06.540

1

you have to defined the variable

 $i=1;
    foreach ($terms as $each) {



             if ($i == 1)
                 $query .= "keywords LIKE '%$each%' ";  
             else
                 $query .= " OR keywords LIKE '%$each%' ";

        $i++;
              }

edited Oct 09 '15 at 04:53

answered Oct 09 '15 at 04:35

Qaisar Satti

2,742
2
18
36

score 1 · Answer 2 · answered Oct 09 '15 at 05:01

1

Before use define i variable value or try this

foreach ($terms as $key => $value) {        
if ($key == 1)
$query .= "keywords LIKE '%$value%' ";  
else
$query .= " OR keywords LIKE '%$value%' ";
}

answered Oct 09 '15 at 05:01

Abhishek Sharma

6,689
1
14
20

score 0 · Answer 3 · answered Oct 09 '15 at 04:35

First define i variable then use like this

$i = 1;         
foreach ($terms as $each) {        
if ($i == 1)
             $query .= "keywords LIKE '%$each%' ";  
         else
             $query .= " OR keywords LIKE '%$each%' ";
          $i++;
     }

I am getting Notice: Undefined variable: i in C:\xampp\htdocs\search-engine\search\search.php on line 23

3 Answers3