How to take value from array into integer (PHP)

Question

Input :

$sql1 = "SELECT COUNT(*) FROM matchTrip where userTripId = :tripId";

$stmt1 = $this->db->prepare($sql1);

$stmt1->bindParam(':tripId', $trip, PDO::PARAM_INT);                                              

$temp = $stmt1->fetchObject();

echo(json_encode($temp));

Output:

 How to take value from array : 

 of which json_encode looks like this: {"COUNT(*)":"7"}

Any help will be appreciated.

@Gautam3164 : when I print only $temp then it says print only this : Array — Ponting
– Ponting, Commented Sep 2, 2013 at 9:20

Fluffeh · Accepted Answer · 2013-09-02 09:20:18Z

1

Why don't you just give the column an alias in the SQL itself?

$sql1 = "SELECT COUNT(*) as myCount FROM matchTrip where userTripId = :tripId";

Makes the rest easier to work with.

answered Sep 2, 2013 at 9:20

Fluffeh

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2 Comments

Ponting Over a year ago

Ok.But how to print myCount?

Ponting Over a year ago

Got it.$temp->myCount.Thanks

Gerben Jacobs · Accepted Answer · 2013-09-02 09:16:51Z

0

Do you mean json_decode? You can just put it between quotes and it should work; $array["COUNT(*)"].

But you can also add "AS myCount" to your SQL.

answered Sep 2, 2013 at 9:16

Gerben Jacobs

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Comments

Your Common Sense · Accepted Answer · 2013-09-02 09:20:27Z

0

if to get rid of all the useless stuff from your code

$sql  = "SELECT COUNT(*) FROM matchTrip where userTripId = ?";
$stmt = $this->db->prepare($sql);
$stmt->execute(array($table_of_user[$i]));
$count = $stmt->fetchColumn();

echo $count;

answered Sep 2, 2013 at 9:20

Your Common Sense

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Comments

xdazz · Accepted Answer · 2013-09-02 09:20:30Z

0

So why not just fecth as array?

$temp = $stmt1->fetch(PDO::FETCH_ASSOC);
echo $temp['COUNT(*)'];

answered Sep 2, 2013 at 9:20

xdazz

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Comments

Bora · Accepted Answer · 2013-09-02 09:22:18Z

0

Just use like this:

$json = json_encode($temp);

echo $json->{'COUNT(*)'}; // 7

answered Sep 2, 2013 at 9:22

Bora

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