How do I modify this MySQL INSERT Query for multiple rows? - php

Question

I am attempting to insert records for Artists, Songs and Record Labels, whilst checking that the data does not already exist in the Database.
The following code is from Mike Fenwick.
<?php
$query = "SELECT id FROM table WHERE unique1=value1 AND unique2=value2…";
$select_result = mysql_query($query);
if (!mysql_num_rows($select_result)) {
$query = "INSERT INTO table SET unique1=value1 AND unique2=value2…";
$insert_result = mysql_query($query);
$id = mysql_insert_id();
}
else {
$row = mysql_fetch_assoc($select_result);
$id = $row['id'];
}
return $id;
?>
I need to modify this to check if three unique values exist already (from 3 separate tables), and if not, insert them. Here is my attempt:
<?php
$query = "SELECT id FROM artistsTable WHERE artistName='Beyonce'";
$select_result = mysql_query($query);
if (!mysql_num_rows($select_result)) {
$query = "INSERT INTO table SET artistName='Beyonce' AND artistImage='beyonce.jpg'";
$insert_result = mysql_query($query);
$artistID = mysql_insert_id();
}
else {
$row = mysql_fetch_assoc($select_result);
$artistID = $row['artistID'];
}
return $artistID;
$query = "SELECT id FROM recordLabelTable WHERE labelName='Columbia Records'";
$select_result = mysql_query($query);
if (!mysql_num_rows($select_result)) {
$query = "INSERT INTO table SET labelName='Columbia Records'";
$insert_result = mysql_query($query);
$labelID = mysql_insert_id();
}
else {
$row = mysql_fetch_assoc($select_result);
$labelID = $row['labelID'];
}
return $labelID;
$query = "SELECT id FROM songTable WHERE trackTitle='Crazy in Love' AND artistID=".$artistID." AND labelID=".$labelID."";
$select_result = mysql_query($query);
if (!mysql_num_rows($select_result)) {
$query = "INSERT INTO songTable SET trackTitle='Crazy in Love' AND artistID=".$artistID." AND labelID=".$labelID."";
$insert_result = mysql_query($query);
$songID = mysql_insert_id();
}
else {
$row = mysql_fetch_assoc($select_result);
$songID = $row['songID'];
}
return $songID;
?>
I'm assuming that there must be a more efficient way to do this. Any ideas would be much appreciated.

Answer 1

Using basic inset / ignore syntax you could do something like this.
A couple of inserts to put in the artist details and label details, then an INSERT based on a SELECT:-
<?php
$query = "INSERT IGNORE INTO artistTable (artistName, artistImag) VALUES('Beyonce', 'beyonce.jpg')";
$insert_result = mysql_query($query);
$query = "INSERT IGNORE INTO labelTable (labelName) VALUES('Columbia Records')";
$insert_result = mysql_query($query);
$query = "INSERT IGNORE INTO songTable (trackTitle, artistID, labelID)
SELECT 'Crazy in Love', a.artistID, b.labelID
FROM artistTable a
INNER JOIN labelTable b
ON a.artistName = 'Beyonce'
AND a.artistImag = 'beyonce.jpg'
AND b.labelName = 'Columbia Records'";
$insert_result = mysql_query($query);
$songID = mysql_insert_id();
return $songID;
?>
As #LoganWayne says, you probably should use MySQLi .

Answer 2

<?php
/* ESTABLISH CONNECTION */
$con=mysqli_connect("Host","Username","Password","Database"); /* REPLACE NECESSARY DATA */
if(mysqli_connect_errno()){
echo "Error".mysqli_connect_error();
}
/* FOR artistsTable TABLE */
$query = "SELECT id FROM artistsTable WHERE artistName='Beyonce'";
$select_result = mysqli_query($con,$query); /* EXECUTE QUERY */
if (mysqli_num_rows($select_result)==0) { /* IF QUERY'S RESULT IS 0 */
$query = "INSERT INTO table SET artistName='Beyonce' AND artistImage='beyonce.jpg'";
$insert_result = mysqli_query($con,$query); /* EXECUTE INSERT QUERY */
} /* END OF IF */
else {
while($row = mysqli_fetch_array($select_result)){
$artistID = mysqli_real_escape_string($con,$row['artistID']); /* ESCAPE STRING */
} /* END OF WHILE LOOP */
} /* END OF ELSE */
/* FOR recordLabelTable TABLE */
$query = "SELECT id FROM recordLabelTable WHERE labelName='Columbia Records'";
$select_result = mysqli_query($con,$query); /* EXECUTE SELECT QUERY */
if (mysqli_num_rows($select_result)==0) { /* IF QUERY'S RESULT IS 0 */
$query = "INSERT INTO table SET labelName='Columbia Records'";
$insert_result = mysqli_query($con,$query); /* EXECUTE INSERT QUERY */
}
else {
while($row = mysqli_fetch_array($select_result)){
$labelID = mysqli_real_escape_string($con,$row['labelID']); /* ESCAPE STRING */
}
}
/* FOR songtable TABLE */
$query = "SELECT id FROM songTable WHERE trackTitle='Crazy in Love' AND artistID='$artistID' AND labelID='$labelID'";
$select_result = mysqli_query($con,$query); /* EXECUTE SELECT QUERY */
if (mysqli_num_rows($select_result)==0) {
$query = "INSERT INTO songTable SET trackTitle='Crazy in Love' AND artistID='$artistID' AND labelID='$labelID'";
$insert_result = mysqli_query($con,$query); /* EXECUTE QUERY */
} /* END OF IF */
else {
while($row = mysqli_fetch_array($select_result)){
$songID = mysqli_real_escape_string($con,$row['songID']);
} /* END OF WHILE LOOP */
}
?>
SUMMARY:
Used at least MySQLi instead of deprecated MySQL.
Replaced fetch_assoc() function with fetch_array() function.
Used mysqli_real_escape_string() function to prevent some of SQL injections.
Cleaned your code. You have misplaced apostrophes(') and double quotes(") hanging around.