PHP string is not joining together

Question

PHP code:

    $checked = $this->input->post('checkbox');
    $status = $this->input->post('adsstatus');
    if(in_array('all', $checked)){
        $query = 'SELECT * FROM ads'; 
    }
    else{
        $finalnumber = count($checked) - 1;
        for($i = 0; $i < count($checked); $i++){
            $query = 'SELECT * FROM ads WHERE ';
            if($i == $finalnumber){
               $query .= 'ID = $checked[$i] '; 
            }
            else{
                $query .= 'ID = $checked[$i] || '; 
            }
        }
    }
    print_r($checked);
    echo $query;

OUTPUT:

Array
(
    [0] => electronics
    [1] => kitchen
    [2] => decoratives
    [3] => homedecor
    [4] => furnitures
    [5] => toys
    [6] => vehicles
)
SELECT * FROM ads WHERE ID = $checked[$i]

the string of variable $query is not joining together. the value of the variable checked is 7 it executing my nested else condition but still string is not joining together. How to resolve it please help me right now I am a helpless

It should be `"ID = {$checked[$i]} "`. You need to use double quotes, and put `{}` around the variable. — Barmar, Oct 13 '16 at 05:15
@Barmar I am a talking about a string is not joining together. — shah rushabh, Oct 13 '16 at 05:16
It is joining together. It's just not replacing `$checked[$i]` with its value, because you used the wrong type of quotes. Variables are only replaced inside double quotes, not single quotes. Didn't you read the answers to the question I linked to? — Barmar, Oct 13 '16 at 05:17
You shouldn't be creating queries by joining strings together anyway. You should use a prepared query with `bind_param()`. — Barmar, Oct 13 '16 at 05:18
Edit your question and add the new version of your code at the end. — Barmar, Oct 13 '16 at 05:21
BTW, you should learn to use `foreach()`, it makes looping over arrays much simpler. — Barmar, Oct 13 '16 at 05:22
@Barmar SELECT * FROM ads WHERE ID = toys it is my latest output. — shah rushabh, Oct 13 '16 at 05:22
You need quotes around the value, since it's a MYSQL string. `$query .= "ID = '{$checked[$i]}'";` — Barmar, Oct 13 '16 at 05:23
Also, instead of using `ID = x OR ID = y OR ...` you should use `WHERE ID IN (x, y, ...)`. — Barmar, Oct 13 '16 at 05:24
Edit the question and show your new code. Otherwise I can't help you.\ — Barmar, Oct 13 '16 at 05:33

PHP string is not joining together

0 Answers0