Multiple PHP mysqli prepared statements (INSERT UPDATE SELECT) weird behavior [duplicate] - php

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What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(1 answer)
Closed 6 years ago.
I have a problem using following PHP function:
function saveCommentOnDB($arg_textComment, $arg_score, $arg_userEmail)
{
$result_tmp = null;
$this->conn->autocommit(false);
echo "saving\n";
echo "text comment: \n";
var_dump($arg_textComment); // OKAY
echo "comment score: \n";
var_dump($arg_score); // OKAY
echo "user mail: \n";
var_dump($arg_userEmail); // OKAY
try {
//[tag1] $query_1 = "INSERT INTO commenti (userFirstname, userEmail, textComment, score, feedback) VALUES ( (SELECT firstname FROM utente u WHERE u.userEmail = 'asd#asd.asd') ,'asd#asd.asd', 'This is an example comment.', 5, 0);";
$query_1 = "INSERT INTO commenti (userFirstname, userEmail, textComment, score, feedback) VALUES ( (SELECT firstname FROM utente u WHERE u.userEmail = ?) ,?,?, ?, 0);";
$query_2 = "UPDATE utente SET commentID=(SELECT c.commentID FROM commenti c WHERE c.userEmail = ?) WHERE userEmail = ?;";
$query_3 = "SELECT commentID, textComment FROM commenti WHERE userEmail = ?;";
$stmt1 = $this->conn->prepare($query_1);
$stmt2 = $this->conn->prepare($query_2);
$stmt3 = $this->conn->prepare($query_3);
$stmt1->bind_param("sssd", $arg_userEmail, $arg_userEmail, $arg_textComment, $arg_score);
$stmt2->bind_param("ss", $arg_userEmail, $arg_userEmail);
$stmt3->bind_param("s", $arg_userEmail);
$stmt1->execute();
$stmt2->execute();
$stmt3->execute();
$stmt3->bind_result($col1, $col2);
$stmt3->fetch();
echo "result:\n";
var_dump($col1); // OKAY
var_dump($col2); // OKAY
$result_tmp = array(
'commentID' => $col1,
'textComment' => $col2
);
$this->conn->commit();
} catch (Exception $e) {
$this->conn->rollback();
}
return $result_tmp;
}
Please, ignore the echo and var_dump, I put them only for debugging.
The problem is that in this function these three prepared statement seems to work not correctly. In particular the statement $stmt1: the result of $stmt3 is correct (as if $stmt1 and $stmt2 are executed correctly), but I don't see anything on my Database. In other words: the statements works correctly 'temporarily' during the execution, but in MyPHP Admin there's nothing on the table commenti.
For example, we assume having this on the DB:
Now I launch the function with following parameters:
$arg_textComment = 'This is an example comment'
$arg_score = '5'
$arg_userEmail = 'asd#asd.asd'
and we have on my browser console:
ie: the commentID (28) is right and the comment text (commentcomment) was "saved", then I recheck the DB but I have still this:
and var_dump($stmt1) after the execution is:
stmt1:
object(mysqli_stmt)#4 (10) {
["affected_rows"]=>
int(1)
["insert_id"]=>
int(41)
["num_rows"]=>
int(0)
["param_count"]=>
int(4)
["field_count"]=>
int(0)
["errno"]=>
int(0)
["error"]=>
string(0) ""
["error_list"]=>
array(0) {
}
["sqlstate"]=>
string(5) "00000"
["id"]=>
int(4)
}
The var_dump seems to be ok, but DB nope.
So I try to execute the query 'manually' by this (it will be executed only the code into the green lined box):
and I have what I expected:
sql> INSERT INTO commenti (userFirstname, userEmail, textComment, score, feedback) VALUES ( (SELECT firstname FROM utente u WHERE u.userEmail = 'asd#asd.asd') ,'asd#asd.asd', 'commentcomment', '5', 0) [2017-01-21 17:38:28] 1 row affected in 11ms
Keep in mind score value is store on DB as float.
The SQL query of the $stmt1 is the same I inserted manually (INSERT INTO... via PHPStorm).
Why the first doesn't works and instead the second yes?
Hope this screencast may help:
https://youtu.be/UsYK93jYVqA

Answer 1

Problem solved, change from this:
$stmt1->execute();
$stmt2->execute();
$stmt3->execute();
to this:
$stmt1->execute();
$this->conn->commit();
$stmt2->execute();
$this->conn->commit();
$stmt3->execute();
$this->conn->commit();
Have no idea of why... 😐 but it works after many tests.