Error: Array to string conversion error

Question

Firstly I am randomly selecting the ID from my table. That part works fine but the next part doesn't. The next part is selecting the ID's row, e.g. if the ID is 6, then it should select all the fields related to 6.

my table is like this:

------------------------------
|ID|Name|Email      |Password|
------------------------------
|1 |Amy |[email protected]|jaaaaaaa|
------------------------------
|2 |Bob |[email protected]|haaukanm|
------------------------------
|3 |Bill|[email protected]|fsoji443|
------------------------------

This is my code:

<?php
include('connect.php');

//select a number between min id and max id
$v = "SELECT ID FROM `tblaccounts` ORDER BY RAND() LIMIT 1";
$result = mysqli_query($connection, $v);
$data2 = mysqli_fetch_array($result);
//var_dump($data2);

$c = "SELECT * FROM `tblaccounts` WHERE ID='$data2'";
$cresult = mysqli_query($connection, $c);
$data3 = mysqli_fetch_array($cresult);
var_dump($data3);
?>

Mayank Pandeyz · Accepted Answer · 2017-06-29 07:55:15Z

4

The issue is here:

$c = "SELECT * FROM `tblaccounts` WHERE ID='$data2'";

here $data2 is not a single value, its an array and you are trying to compare it in WHERE like a string, that's why the error. Instead try $data2['id'] like:

$c = "SELECT * FROM `tblaccounts` WHERE ID='".$data2['ID']."'";

or

$c = "SELECT * FROM `tblaccounts` WHERE ID=".$data2['ID'];  // Sinlge quote is not required if `ID` is `int`

answered Jun 29, 2017 at 7:55

Mayank Pandeyz

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Comments

Viet Nguyen · Accepted Answer · 2017-06-29 07:57:12Z

1

Because your $data2 is an array, this is should work

include('connect.php');

//select a number between min id and max id
$v = "SELECT ID FROM `tblaccounts` ORDER BY RAND() LIMIT 1";
$result = mysqli_query($connection, $v);
$data2 = mysqli_fetch_array($result);

//var_dump($data2);
$c = "SELECT * FROM `tblaccounts` WHERE ID='".$data2['ID']."'";
$cresult = mysqli_query($connection, $c);
$data3 = mysqli_fetch_array($cresult);
var_dump($data3);

answered Jun 29, 2017 at 7:57

Viet Nguyen

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Comments

Passionate Coder · Accepted Answer · 2017-06-29 08:13:44Z

0

You are getting this error as you are comparing array in where clasue. Your $data is an array like below

$data = array(
             'ID'=>2
             'Name'=>'Bob',
             'Email'=>'[email protected]',
             'Password'=>'haaukanm'

       );// say record with id 2 is fecthed

So use $data['ID'] in your where clause

answered Jun 29, 2017 at 8:13

Passionate Coder

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