I am learning PHP and SQL, and I'm trying to figure out how to select a record from a database.
I created a function called selectById()
Right now in the browser displayed is "Error:" but, no specific error was displayed.
// function selectById --------------------------------------------------------------------
function selectById($pUInput) {
$sql = mysql_query("SELECT * FROM tblStudents
WHERE id='$pUInput[0]'");
if (!mysql_query($sql))
{
die('Error: ' . mysql_error());
}
echo "Record Selected";
}
PHP Code:
//Call function mainline
mainline();
// Declare the function mainline
function mainline() {
$uInput = getUserInput();
$connectDb = openConnect(); // Open Database Connection
selectDb($connectDb); // Select Database
doAction($uInput);
//display();
//closeConnect();
}
//Declare function getUserInput --------------------------------------------------------
function getUserInput() {
echo "In the function getUserInput()" . "<br/>";
// Variables of User Input
$idnum = $_POST["idnum"]; // id (NOTE: auto increments in database)
$fname = $_POST["fname"]; // first name
$lname = $_POST["lname"]; // last name
$major = $_POST["major"]; // major
$year = $_POST["year"]; // year
$action = $_POST["action"]; // action (select, insert, update, delete)
$userInput = array($idnum, $fname, $lname, $major, $year, $action);
return $userInput;
}
function doAction($pUserInput) {
echo "In function doAction()" . "<br/>";
if ($pUserInput[5] == "sel") {
selectById($pUserInput);
} elseif ($pUserInput[5] == "ins") {
insert($pUserInput);
}
}
// Create a database connection -----------------开发者_开发百科---------------------------------------
function openConnect() {
$connection = mysql_connect("localhost", "root_user", "password");
echo "Opened Connection!" . "<br/>";
if(!$connection) {
die("Database connection failed: " . mysql_error());
}
return $connection;
}
// Select a database to ----------------------------------------------------------------
function selectDb($pConnectDb) {
$dbSelect = mysql_select_db("School", $pConnectDb);
if(!$dbSelect) {
die("Database selection failed: " . mysql_error());
} else {
echo "You are in the School database! <br/>";
}
}
// function select ---------------------------------------------------------------------
function selectById($pUInput) {
$sql = mysql_query("SELECT * FROM tblStudents
WHERE id='$pUInput[0]'");
if (!mysql_query($sql))
{
die('Error: ' . mysql_error());
}
echo "Record Selected";
}
// function insert -----------------------------------------------------------------------------
function insert($pUInput) {
$sql="INSERT INTO tblStudents (first_name, last_name, major, year)
VALUES
('$pUInput[1]','$pUInput[2]','$pUInput[3]', '$pUInput[4]')";
if (!mysql_query($sql))
{
die('Error: ' . mysql_error());
}
echo "1 record added";
}
?>
SQL Syntax:
CREATE TABLE `tblStudents` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`first_name` varchar(30) NOT NULL,
`last_name` varchar(50) NOT NULL,
`major` varchar(40) NOT NULL,
`year` date NOT NULL,
PRIMARY KEY (`id`)
)
You are running a query on a query result. This will not work. You will need to use something along the lines of
$sql = mysql_query("SELECT * FROM tblStudents
WHERE id='" . $pUInput[0] . "'");
if (!$row = mysql_fetch_assoc($sql))
Which would assign $row an array value if the query did not fail. You may also want to filter the pUinput as well with mysql_real_escape_string as you do not necessarily know what it contains (or statically cast it to an integer).
EDIT
Added a bit extra information.
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