Array to string conversion failed in PHP

Question

I have a query which fetches distinct user ids and I am trying to convert it into a comma separated string which I can pass into another sql query having IN in WHERE clause. But I am getting an error saying array to string conversion.

$qry0="SELECT DISTINCT id FROM users  ";
$res0=getData($qry0);
while($row0=mysqli_fetch_array($res0))
{
    $data0      =   $row0['id'];  
}

And I'm trying to convert it as string like this:

$array = explode(",", $data0);

and pass it to another

$qry="SELECT * FROM login WHERE clientid IN(".$array.") ";

explode creates an array. where implode creates a string ... an SQL query requires a string, not an array. — Peon
– Peon, Commented May 19, 2016 at 13:46
$data0 is not an array.It is a variable with last value of id from your loop . — Bikash
– Bikash, Commented May 19, 2016 at 13:46

Jayesh Chitroda · Accepted Answer · 2016-05-19 13:46:49Z

1

USe implode instead of explode:

$qry0="SELECT DISTINCT id FROM users  ";
$res0=getData($qry0);
$data0 = array(); // initialize array first
while($row0=mysqli_fetch_array($res0))
{
    $data0[]      =   $row0['id'];  // create array like this
}

$array = implode(",", $data0); // use implode to convert array to string

answered May 19, 2016 at 13:46

Jayesh Chitroda

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Comments

Ranjit Shinde · Accepted Answer · 2016-05-19 14:00:34Z

0

The explode() function breaks a string into an array.To break array into string you need to use implode()

$qry0="SELECT DISTINCT id FROM users  ";
$res0=getData($qry0);
$data0 = array();
while($row0=mysqli_fetch_array($res0))
{
    $data0[]      =   $row0['id'];  
}
$array = implode(",", $data0);

edited May 19, 2016 at 14:00

answered May 19, 2016 at 13:46

Ranjit Shinde

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