Inserting a row into a table using PHP MYSQL - php

Question

The following php code is meant to drop a table if it exists, create the table, use the table, and then insert a row into the table.
Everything works apart from the insert. I am new to PHP and MYSQL and I have tried many permutations of different types of quotes (singles, doubles, this one: `) but cannot get the data to be inserted into the table.
Can anybody shed some light on what is wrong with this?
$retval = mysqli_query($conn,'INSERT INTO `performance` (manager, program, programid, yearmonth, performance) VALUES ("manager1", "program1","programid1", "199901", "-3.4")');
The php script below gives the output:
Connected successfully
Table dropped successfully.
DB used successfully.
Table created successfully.
Could not insert data.
So everything worked apart from the insert.
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully<br />';
$retval = mysql_query('DROP TABLE IF EXISTS `managedfutures`.`performance`') or die(mysql_error());
if(! $retval )
{
die('Could not drop table ' . mysql_error());
}
echo "Table dropped successfully.";
echo "<br>";
$retval = mysql_query("USE managedfutures", $conn);
if(! $retval )
{
die('Could not use DB' . mysql_error());
}
echo "DB used successfully.";
echo "<br>";
$sql = "CREATE TABLE performance( ".
"performance_id INT NOT NULL AUTO_INCREMENT, ".
"manager VARCHAR(255) NOT NULL, ".
"program VARCHAR(255) NOT NULL, ".
"programid VARCHAR(255) NOT NULL, ".
"yearmonth VARCHAR(6) NOT NULL, ".
"performance FLOAT NOT NULL, ".
"PRIMARY KEY (performance_id )); ";
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Could not create table: ' . mysql_error());
}
echo "Table created successfully.";
echo "<br>";
$retval = mysqli_query($conn,'INSERT INTO `performance` (manager, program, programid, yearmonth, performance) VALUES ("manager1", "program1","programid1", "199901", "-3.4")');
if(! $retval )
{
die('Could not insert data. ' . mysql_error());
}
echo "Data inserted successfully.";
echo "<br>";
return;
Thanks to Mike W for pointing out that I had mixed mysql and mysqli commands! I am new to php/mysql and did not realise that there was a difference between the two. There was another error also, I was inputting a number as a string in the insert statement. I.e. I wrote "-3.4" instead of just -3.4.
For completeness, here is the fixed version which works.
$mysqli = new mysqli($dbhost, $dbuser, $dbpass, $dbname);
if ($mysqli->connect_errno) {
die("Failed to connect to MySQL: " . $mysqli->connect_error);
}
echo 'Connected successfully<br />';
$retval = mysqli_query($mysqli,"DROP TABLE IF EXISTS `performance`");
if(! $retval )
{
die('Could not drop table ' . $mysqli->query_error);
}
echo "Table dropped successfully.";
echo "<br>";
$sql = "CREATE TABLE performance( ".
"performance_id INT NOT NULL AUTO_INCREMENT, ".
"manager VARCHAR(255) NOT NULL, ".
"program VARCHAR(255) NOT NULL, ".
"programid VARCHAR(255) NOT NULL, ".
"yearmonth VARCHAR(6) NOT NULL, ".
"performance FLOAT NOT NULL, ".
"PRIMARY KEY (performance_id )); ";
$retval = mysqli_query($mysqli, $sql);
if(! $retval )
{
die('Could not create table: ' . $mysqli->query_error);
}
echo "Table created successfully.";
echo "<br>";
$retval = mysqli_query($mysqli, "INSERT INTO `performance` (`manager`, `program`,`programid`, `yearmonth`, `performance`) VALUES ('manager1', 'program1','programid1', '199901', -3.4)");
if(! $retval )
{
die('Could not insert data. ' . $mysqli->query_error);
}
echo "Data inserted successfully.";
echo "<br>";
return;

Answer 1

You're mixing mysql_*() and mysqli_*() calls. The two are different and cannot be used together. mysql_*() is deprecated - use only mysqli_*().

Answer 2

You have used mysql_query throughout your code
$retval = mysql_query( $sql, $conn ); //**You have used mysql_query**
if(! $retval )
{
die('Could not create table: ' . mysql_error());
}
echo "Table created successfully.";
echo "<br>";
Suddenly a mysqli_query is seen (MAGIC !!!).
$retval = mysqli_query($conn,'INSERT INTO `performance` (manager, program, programid, yearmonth, performance) VALUES ("manager1", "program1","programid1", "199901", "-3.4")');
^^
// SUDDENLY you see mysqli_query