Return "alert" if an IF is not executed

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Return "alert" if an IF is not executed

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#1 by (1 votes)

1

How can I check if an if was executed and if it did not display an alert?

See the example:

if ($newtime!=0) {
   $qryupd = "update registration set dados=" . $dadosnew . " where id=$id";
   $result2 = mysqli_query($db, $qryupd);
   if (!$result2) {
      rollback();
      echo '[{"resultado":"falhou","total_lances":"' . $bal . '","motivo":"Falha ao executar lance, por favor tente novamente! Código: LW102"}]';
      exit;
   }
}

In this example it already does this for the query that updates in the DB, if it does not update it shows the alert. I need to do the same thing, but I need it to check the first IF condition, in the case: if ($newtime!=0) {

Is it possible to do this?

php mysql

asked by anonymous 21.11.2017 / 02:20

1 answer

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score 1 · Accepted Answer

            if ($newtime!=0) {
        $qryupd = "update registration set dados=" . $dadosnew . " where id=$id";
        $result2 = mysqli_query($db, $qryupd);
        if (!$result2) {
            rollback();
            echo '[{"resultado":"falhou","total_lances":"' . $bal . '","motivo":"Falha ao executar lance, por favor tente novamente! Código: LW102"}]';
            exit;
        }
        }else{
    echo "newtime é igual a 0";
    echo "<script>alert('VARIAVEL $NEWTIME É IGUAL A 0')</script>;

}

I think this will solve your problem. You can use the IF / ELSE.

if(condicao)
{
entrou na condição
}else{
não entrou na condição
}