I have a PHP script which fires a MySQL SP. The SP cmd works when run directly against db (I'm using PhpMyAdmin UI). When tested, the PHP returns error 'Warning: mysql_num_rows() expects parameter 1 to be resource' due to the fact there is nothing returned from the SP...when i debug the script i see the variable $result1 has all its params set to null.
There is another SP executed successfully using the same code above this one in the script so im a bit puzzled. Any ideas what the issue is?
$result1 = $mysqli->query("CALL sp_playerInvite_pin_select('samsmith#hotmail.com')");
if(!$result1) die("CALL failed: (" . $mysqli->errno . ") " . $mysqli->error);
//echo json_encode($getPlayerInvitePin);
// check for empty result, if not then there is >1 pin corresponding to email
if (mysql_num_rows($result1) > 0)
{
//looping through all results-in theory should only 1 result
while ($row = mysql_fetch_array($result1))
{
$response["pin"] = $row["player_pin"];
}
// echoing JSON response
echo json_encode($response);
}
The SP...
Shouldn't it be mysqli_num_rows? thanks 'I Can Has Cheezburger' :
Related
I have a web page created in php using html code. I want to save user information entered in my web page to a MySQL database. I am using php as the middle man to link the frontend web page(htmnl code) to the database(mysql).
Inside my link folder (middle man php file) I have the following:
<?php
//Gets server connection credentials stored in serConCred2.php
require_once('ConCred2.php');
//SQL code for connection w/ error control
$con = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
if(!$con){
die('Could not connect: ' . mysqli_connect_error());
}
//Selection of the databse w/ error control
$db_selected = mysqli_select_db($con, DB_NAME);
if(!$db_selected){
die('Can not use ' . DB_NAME . ': ' . mysqli_error($con));
}
//Co-PI and Co-Investigator Information variables
$Co_FNAME = $_POST['fname'];
$Co_LNAME = $_POST['lname'];
$Co_SLNAME = $_POST['slname'];
$Co_DEGREE = $_POST['Degree_Selection'];
$Co_DEGREE_Other = $_POST['other_specify_degree']; //hold the value of degree if user selected other from the dropdown menu
$Co_CPOS = $_POST['Current_Position_Selection'];
$Co_CPOS_Other = $_POST['other_specify_cpos']; //hold the value of Current Position if user selected other from the dropdown menu
$Co_INST = $_POST['Institution_Selection'];
$Co_INST_Other = $_POST['other_specify_inst']; //hold the value of Current Position if user selected other from the dropdown menu
$Co_SCHOOL = $_POST['School_Selection'];
$Co_SCHOOL_Other = $_POST['other_specify_school']; //hold the value of Current Position if user selected other from the dropdown menu
$Co_DEPART = $_POST['Department_Selection']; //Este se estara eliminando en la version online
$Co_DEPART_Other = $_POST['other_specify_department']; //hold the value of Department if user selected other from the dropdown menu
$Co_PROGRAM = $_POST['program'];
$Co_EMAIL = $_POST['email'];
$Co_PHONE = $_POST['phone'];
//If decition when user select other from the dropdown menu
if($Co_DEGREE == "other_degree") $Co_DEGREE = $Co_DEGREE_Other;
if($Co_CPOS == "other_cpos") $Co_CPOS = $Co_CPOS_Other;
if($Co_INST == "other_inst") $Co_INST = $Co_INST_Other;
if($Co_SCHOOL == "other_school") $Co_SCHOOL = $Co_SCHOOL_Other;
if($Co_DEPART_Other == "other_department") $Co_DEPART = $Co_DEPART_Other;
//This sets a starting point in the rollback process in case of errors along the code
$success = true; //Flag to determine success of transaction
//start transaction
echo "<br>1. Going to set autocommit to 0";
$command = "SET AUTOCOMMIT = 0";
echo "<br>2. Autocomint has been set to 0";
echo "<br>3. Going to run query to see if result is true or false";
$result = mysqli_query($con, $command);
echo "<br>4. Finished running the query. Result is:" . $result;
echo "<br>5. Going to set command to BEGIN";
$command = "BEGIN";
echo "<br>6. Command is now BEGIN";
echo "<br>7. Going to run query for command BEGIN";
$result = mysqli_query($con, $command);
echo "<br>8. Query runned for command BEGIN";
echo "<br>9. Result value is: " . $result;
//Saves Pi values into database
/**
$sqlCoPI = "INSERT INTO co_pi_table (Fname, Lname, SLname, Degree, Current_Position, Institution, School, Department, Program, Email, Phone)
VALUES('$Co_FNAME', '$Co_LNAME', '$Co_SLNAME', '$Co_DEGREE', '$Co_CPOS', '$Co_INST', '$Co_SCHOOL', '$Co_DEPART', '$Co_PROGRAM', '$Co_EMAIL', '$Co_PHONE')";
*/
echo "<br>10. Going to write sql command to populate table pi_table";
/**
$sqlPi = "INSERT INTO pi_table (Fname, Lname, SLname, Degree, Current_Position, Institution, School, Department, Program, Email, Phone)
VALUES('$Co_FNAME', '$Co_LNAME', '$Co_SLNAME', '$Co_DEGREE', '$Co_CPOS', '$Co_INST', '$Co_SCHOOL', '$Co_DEPART', '$Co_PROGRAM', '$Co_EMAIL', '$Co_PHONE')";
*/
$sqlPi = "INSERT INTO pi_table (Fname) VALUES('$Co_FNAME')";
//Checks to see if theres an error in the pi db con
echo "<br>11. Sql command finished writting.";
echo "<br>12. Going to query the sql finished command to the database to determine value of result.";
$result = mysqli_query($con, $sqlPi);
echo "<br>13. Finished running sql command to database. Result value is: " . $result;
echo "<br>14. Going to enter if statements depending on result value";
if($result == false){
//die ('<br>Error in query to PI table: ' . mysqli_error($con));
echo "<br>15. I am inside the false statement. Success is going to be set as false. ";
$success = false;
//$success = true; //Cahnged this in order to test if values are being saved to db. Change back to false.
}
//Checks for errors or craches inside the code
// If found, execute rollback
echo "<br>16. Going to verify is success is true.";
if($success){
$command = "COMMIT";
$result = mysqli_query($con, $command);
//echo "<br>Tables have been saved with 0 errors.";
echo "<br><p style=\"color: red;\"Principal Investigator has been saved successfuly. <br><br>
You may now CLOSE this page and press the<br><br> \"Refresh List\" <br><br>
button to display name in dropdown menu selection.</p>";
}
else{
$command = "ROLLBACK";
$result = mysqli_query($con, $command);
echo "<br>17. Success was determined to be false.";
echo "<br>Error! Databases could not be saved.<br>
Contact system manager to report error. <br> <br>" . mysqli_error($con);
}
echo "<br>18. Setting autocommit back to 1 again.";
$command = "SET AUTOCOMMIT = 1"; //return to autocommit
$result = mysqli_query($con, $command);
//Displays message
//echo '<br>Connection Successfully. ';
//echo '<br>Database have been saved';
//Close the sql connection to dababase
mysqli_close($con)
?>
As you can read, I am requiring users to fill out their information. Some of the information required are dropdown menu fields that user selects an option from among the presented ones.
The problem I am having is, when the above php code executes, it determines that the $result variable is false and doesn't save anything. When you execute the code, you get the following messages displayed:
1. Going to set autocommit to 0
2. Autocomint has been set to 0
3. Going to run query to see if result is true or false
4. Finished running the query. Result is:1
5. Going to set command to BEGIN
6. Command is now BEGIN
7. Going to run query for command BEGIN
8. Query runned for command BEGIN
9. Result value is: 1
10. Going to write sql command to populate table pi_table
11. Sql command finished writting.
12. Going to query the sql finished command to the database to determine value of result.
13. Finished running sql command to database. Result value is:
14. Going to enter if statements depending on result value
15. I am inside the false statement. Success is going to be set as false.
16. Going to verify is success is true.
17. Success was determined to be false.
Error! Databases could not be saved.
Contact system manager to report error.
18. Setting autocommit back to 1 again.
For security purposes I cant post the html content since it has sensitive name information nor the databases. Although I can ensure that the tables inside the database are called exactly as mentioned in the sql command line.
I HAVE FOUND THE PROBLEM!
After long debating I decided to recreate the database In which all the information was being stored. When I redirected the table in my sql command ( Instead of saving it in "pi_table" I saved it in a newly created database called "pi_table_2") and everything worked out properly.
Aparently my database got corrupted and phpMyAdmin didn't recognized that it was curropted.
For reference my database tables where in InnoDB format. What might have cause this to happen, who knows but if you ever encounter a similar problem, creating a small testing database and see if it saves. If it does, recreate the table and it might solve your issue like it solved mine.
Once again thank you a lot guys!!!!!
I am looking at the code and everything seems to be in order, could be a syntax error like a missing quotation for example:
//SQL code for connection w/ error control
$con = mysqli_connect("DB_HOST", "DB_USER", "DB_PASSWORD", "DB_NAME");
also
$db_selected = mysqli_select_db($con, "DB_NAME");
or die ("Cant select Database");
}
Hope this help.
Cheers;
Hasan
$query = "SELECT `ip` FROM `banned` WHERE `ip` = '$ip'";
$retval = mysqli_query($conn, $query);
if(!$retval){
die("Could not Execute Query: " . mysqli_error($conn));
} else {
if(mysqli_num_rows($retval) == 0){
echo "test";
} else {
header('Location: http://www.teutonic-development.net/index.php?p=banned');
}
}
when I'm running this code all that's printed out is: "Could not Execute Query:"
I really have absolutely no idea why it's doing this. I'm connecting fine in my init.php file. Which is where this file is.
My other script which just adds a log entry works fine. And if I run my $query in phpmyadmin's sql interpreter it runs perfectly fine (when I replace the $ip part with an actual ip of course)
Any suggestions?
Normally one would say that hey your query failed to execute story finish. But this case is interesting.
Your code is
die("Could not Execute Query: " . mysqli_error($conn));
and your error message is
Could not Execute Query:
Notice even though you have mysqli_error($conn) but there is no mysql error being shown. That confirms 100% that $conn is not properly established (contrary to what you think)
So take a look at your code again and see if $conn is really a mysqli resource and is available to your file in proper variable scope.
I have been reading about the Commands out of sync; you can't run this command now problem for some time now, and see that you cannot have any unread results left, which makes sense to me. However, in the following case, I don't see which results I am missing to free. I have left out the irrelevant things from my PHP and SQL code below.
# Set local variables
$sql = "
SET #STARTDATE = '2014-09-01';
SET #RANK = 0;
";
if (mysqli_multi_query($conn, $sql)) {
# Success: do nothing else
} else {
# Failure: output the error message
echo "Error: " . $sql . "<br>" . $conn->error;
}
# Fetch and store the results
$sql = "
SELECT * FROM MyTable
";
$result = mysqli_query($conn, $sql);
if (!$result) {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
The second query (the if (!$result) block) returns the infamous Commands out of sync error. If I comment out the first part, the second query runs no problem. If I change the first query into only one SET statement instead of two, the second query runs no problem. Therefore, it seems that I have to clear the 'success-flag' of every individual SQL statement from the first part. Is this correct? If so, how shall this be done?
EDIT: indeed it seems you have to flush all results in between. Adding the following line between part 1 and part 2 solves the problem.
while (mysqli_next_result($conn)) {;} // Flush multi_queries
I found this solution in a user comment on the PHP manual: http://nl3.php.net/manual/en/mysqli.multi-query.php
Quite simply, your first query
SET #STARTDATE = '2014-09-01';
SET #RANK = 0;
Will generate 2 result sets and until they have been processed, even though the result will be just a status you cannot continue.
So you need to do something like this :-
if (mysqli_multi_query($conn, $sql)) {
do {
/* unload result set */
if ($result = $mysqli->store_result()) {
// Check status
$result->free();
}
} while ($mysqli->next_result());
} else {
# Failure: output the error message
echo "Error: " . $sql . "<br>" . $conn->error;
}
Of course you should probably check for errors in that loop
How do I check if a MySQL query is successful other than using die()
I'm trying to achieve...
mysql_query($query);
if(success){
//move file
}
else if(fail){
//display error
}
This is the first example in the manual page for mysql_query:
$result = mysql_query('SELECT * WHERE 1=1');
if (!$result) {
die('Invalid query: ' . mysql_error());
}
If you wish to use something other than die, then I'd suggest trigger_error.
You can use mysql_errno() for this too.
$result = mysql_query($query);
if(mysql_errno()){
echo "MySQL error ".mysql_errno().": "
.mysql_error()."\n<br>When executing <br>\n$query\n<br>";
}
If your query failed, you'll receive a FALSE return value. Otherwise you'll receive a resource/TRUE.
$result = mysql_query($query);
if(!$result){
/* check for error, die, etc */
}
Basically as long as it's not false, you're fine. Afterwards, you can continue your code.
if(!$result)
This part of the code actually runs your query.
mysql_query function is used for executing mysql query in php. mysql_query returns false if query execution fails.Alternatively you can try using mysql_error() function
For e.g
$result=mysql_query($sql)
or
die(mysql_error());
In above code snippet if query execution fails then it will terminate the execution and display mysql error while execution of sql query.
put only :
or die(mysqli_error());
after your query
and it will retern the error as echo
example
// "Your Query" means you can put "Select/Update/Delete/Set" queries here
$qfetch = mysqli_fetch_assoc(mysqli_query("your query")) or die(mysqli_error());
if (mysqli_errno()) {
echo 'error' . mysqli_error();
die();
}
if using MySQLi bind_param try to put this line above the query
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
I have a query written as
mysql_query($query,$conn)
or
die(
"A MySQL error has occurred.<br />Your Query: " . $query . "<br /> Error: (" . mysql_errno() . ") " . mysql_error());
echo "You have been entered into our Database!";
This outputs
A MySQL error has occurred.
Your Query: INSERT INTO users (uid, twname, privacy) VALUES (15400743, 'gdhdh', 'accepted')
Error: (0)
so it doesn't list any errors or anything. When I copy/paste that query into the SQL tab of PHPMyAdmin, it runs successfully, and the DB connection isn't throwing anything bad (I know it works because a query works well elsewhere):
function get_db_conn() {
$conn = mysql_connect($GLOBALS['db_ip'], $GLOBALS['db_user'], $GLOBALS['db_pass']);
mysql_select_db($GLOBALS['db_name'], $conn);
return $conn;
}
Any thoughts on what I could fix? I'd really appreciate it.
Error 0 means that no error occurred.
Therefore if the code to output an error is being run when no error occurred you have a logic error in your code surrounding the call to the mysql_query() function.
It's hard to tell from your code what should and shouldn't run under different conditions and where the error may lie.
The following code is logically equivalent to what you are trying to achieve and should work as expected.
This code more clearly separates calling the mysql_query() function from checking the result of calling the function. I have also formatted the code a little to that it displays without any horizontal scrolling, although that's purely optional.
$queryResult = mysql_query($query, $conn);
if ($queryResult === false) {
$errorMessage = "A MySQL error has occurred.<br />"
. "Your Query: ".$query."<br />"
. " Error: (".mysql_errno().") ".mysql_error();
die($errorMessage);
}
I'm not sure if this is exactly what you're looking for, but the argument to the function is called $query, and the variable in your error is $your_query, so the value of $your_query displays in the error, but who knows what is in $query, so if it's NULL, that might cause the error number 0 with no message.