Why variable value is not showing in insert query

Question

I have this code snippet

$taskAssignedCompleteId = $row["TaskAssignCompletionId"];
echo "taskAssignedCompleteId::".$taskAssignedCompleteId. "\n";

if($taskAssignedCompleteId == 0 || $taskAssignedCompleteId = null)
{
    echo "Currently no task assigned to you. If you are not doing a task on PMS and still receving this massage then, contact your team lead";
    return;
}

$sqlInsert = "INSERT INTO emp_task_finished_request (EmpTaskAssignCompletionId, RequestDateTime) VALUES (".$taskAssignedCompleteId.", '$date')";
echo "output : ".$sqlInsert;
$queryInsert = $connPDO->exec($sqlInsert);

    echo "\n output : ".$queryInsert;

Surprisingly $taskAssignedCompleteId value is not showing in query when i echo my $queryInsert varaiable while it is perfectly showing when i directly echo $taskAssignedCompleteId. Why is the problem? it is very strange for me. here is my output

taskAssignedCompleteId::13
sqlInsert full : INSERT INTO emp_task_finished_request (EmpTaskAssignCompletionId, RequestDateTime) VALUES (, '2017-07-05 16:53:45')
 output :

Not much point using PDO and then introducing an [SQL Injection Attack](http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php) vector by not using [prepared parameterized statements](http://php.net/manual/en/mysqli.quickstart.prepared-statements.php) — RiggsFolly, Jul 05 '17 at 11:56
Can you look this `$taskAssignedCompleteId = null` I think there should be `==` — Ayaz Ali Shah, Jul 05 '17 at 11:57
Why do you have $taskAssignedCompleteId = null? It should be "==" — Joey Ezekiel, Jul 05 '17 at 11:57
@RiggsFolly i am waiting to accept answer, i hope you know stack-overflow waiting policy to accept answer — Muhammad Faizan Khan, Jul 05 '17 at 12:05
I do, but as you have only accepted 7 out of your last 30 questions I though I would provide you a gentle reminder — RiggsFolly, Jul 05 '17 at 12:07

score 1 · Accepted Answer · answered Jul 05 '17 at 12:00

1

You need to correct your if statement. $taskAssignedCompleteId = null is not correct. Kindly replace it by following

if($taskAssignedCompleteId == 0 || $taskAssignedCompleteId == null)

answered Jul 05 '17 at 12:00

Ayaz Ali Shah

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JYoThI · Answer 2 · 2017-07-05T12:00:02.120

0

you should use == for comparison operator . = is assignment operator .so it's assigning null to $taskAssignedCompleteId variable .

if($taskAssignedCompleteId == 0 || $taskAssignedCompleteId == null) { .. }

edited Jul 05 '17 at 12:00

answered Jul 05 '17 at 11:58

JYoThI

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1

You should also put this at comment box. – Jaydeep Mor Jul 05 '17 at 11:59
if it's answer . i can post it in answer area . whats wrong with it @JaydeepMor – JYoThI Jul 05 '17 at 12:02

score 0 · Answer 3 · answered Jul 05 '17 at 11:59

Change $taskAssignedCompleteId = null to $taskAssignedCompleteId == null

By doing this, you are assigning taskAssignedCompleteId to a null value. To compare use "==" but you already know that. Just a typo I guess.

Darn, somebody beat me to the answer as I'm typing this.

Why variable value is not showing in insert query

3 Answers3