Cannot cast string to int (php)

Question

I am trying to finish my question of the day script. I have a random number generator (external) which is inserted into a table daily. I take that number and pull the corresponding question from another table.

// Query 1
$query1 = "SELECT $field1 FROM $table1 WHERE id=1";
$result1 = mysqli_query($con1, $query1);

// Query 2
$query2 = "SELECT $field2 FROM $table2 WHERE QNum = $result1";
$result2 = mysqli_query($con2, $query2);

// Display question
while($row = mysqli_fetch_array($result2)) {
    echo $row['question'];
}

$result1 is pulling a random number, lets say its 9. When $result1 is used to pull the question, it doesn't work but when I replace $result1 with number 9, it works. I experimented with syntax and eventually figured it could be a problem with a string vice an integer.

I tried to cast it as an integer but it keeps assigning the value of $result1 to 1. I am at a loss. I don't understand if the string is a 9, how converting it to an integer would change its value.

I feel like I've tried everything after days of experimenting out there but I am sure it is something very simple. Please help.

You forgot to call mysqli_fetch_array on $result. It doesn't contain the value of the field, it contains a mysqli_result object. — Barmar
– Barmar, Commented Jul 31, 2014 at 4:15

Barmar · Accepted Answer · 2014-07-31 04:24:00Z

2

You're right, it's something very simple. You need to fetch the row to get the value:

$row = mysqli_fetch_assoc($result1);
$qnum = $row[$field1];

$query2 = "SELECT $field2 FROM $table2 WHERE QNum = $qnum";

You seem to understand this in general, since you call mysql_fetch_array to get the results of the second query. Why should the first query be any different?

However, you can do all this in one query:

$query2 = "SELECT t2.$field2
           FROM $table1 t1
           JOIN $table2 t2 ON t1.$field1 = t2.QNum
           WHERE t1.id = 1";

edited Jul 31, 2014 at 4:24

answered Jul 31, 2014 at 4:18

Barmar

789k57 gold badges555 silver badges669 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Nimeshka Srimal · Accepted Answer · 2014-07-31 04:34:49Z

0

Have you checked whats returned in $result1 ??

it should be an object which cannot be used inside the second mysql query. First you have to get the value from that query by fetching the query.

$result1 = mysqli_query($con1, $query1);
$row = mysqli_fetch_array($result1, MYSQLI_NUM);
$random_number = $row[0];

this $random_number then can be used in the second query. hope this will fix your error.

answered Jul 31, 2014 at 4:34

Nimeshka Srimal

9,1605 gold badges48 silver badges62 bronze badges

Collectives™ on Stack Overflow

Cannot cast string to int (php)

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related