mysqli getting warning in php for updateing the data

Question

function update($data, $table, $id)
{
    global $conid;
    //echo $id;
        /*Assuming array keys are = to database fileds*/
        if (count($data) > 0) {
            foreach ($data as $key => $value) {

                $value = mysqli_real_escape_string($value); // this is line shows a warning
                $value = "'$value'";
                $updates[] = "$key = $value";
            }
        }
        $implodeArray = implode(', ', $updates);
        $sql = ("UPDATE $table  SET $implodeArray WHERE id=$id");
        mysqli_query($conid, $sql);


}

This is update code for all forms but it gives a warning. warning shows only in that line other works perfectly i didn't find why this happen..

Warning: mysqli_real_escape_string() expects exactly 2 parameters — Shri, Aug 19 '17 at 07:20

score 0 · Answer 1 · answered Aug 19 '17 at 07:21

0

Since your using mysqli_* function then you have to include your connection to database into mysqli_real_escape function :

You need to change

$value = mysqli_real_escape_string($value);

to

$value = mysqli_real_escape_string($conid, $value);

answered Aug 19 '17 at 07:21

Milan Chheda

8,159
3
20
35

ok. Thank you................. – Shri Aug 19 '17 at 07:24
You’re welcome, If it helped you, please [upvote and accept the answer](https://stackoverflow.com/help/someone-answers) Happy Coding! :) – Milan Chheda Aug 19 '17 at 07:25

score 0 · Answer 2 · answered Aug 19 '17 at 07:27

0

mysql

 $value = mysqli_real_escape_string($value);

mysqli //in mysqli 2 perameter needed as connection object and string data

 $value = mysqli_real_escape_string($conid,$value);//connection object

answered Aug 19 '17 at 07:27

Bhargav Chudasama

6,928
5
21
39

mysqli getting warning in php for updateing the data

2 Answers2