if语句未按预期评估

Question

In the code below, the if statement is evaluated as FALSE where I think it should be TRUE. What am I misunderstanding?

$sql = "SELECT * FROM LIVE";     
$res = mysqli_query($con, $sql);
while ($result = mysqli_fetch_assoc ($res)){
var_dump($result['LIVE']); //string '1' (length=1)
echo $result['LIVE']. "<br>"; //1
}
$a = "1";

if($result['LIVE'] == $a){
echo "It is live!";
}
else {
echo "not live!"; //this gets echoed
}

I have tried all variations of quotes ', ", no quotes

AND changing the data and datatype for the column in the MYSQL table to BIT, CHAR, INT

AND CAST in PHP to BOOL and 'integer'

AND changing the operator eg =, ==, === and have also tried just typing into the if() instead of passing in $a as shown

AND $result = $result['LIVE']; and passing in $result.

I have read PHP type comparison tables and PHP type juggling as well as all the other relative parts of the manual if(), while(), mysqli_fetch_array() etc and can't solve this myself.

I am still learning PHP and I am sure it is an easy fix but I just can't see it.

Answer 1

your not saving the value of $result; during the final iteration of while loop $result will be set to false because there are no more rows to traverse. which is also what triggers to exit the while loop.

you need to set your value to another temp variable.

while ($result = mysqli_fetch_assoc ($res)){
$temp = $result['LIVE']; //1
}
$a = "1";

if($temp == $a){