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Php : Take Input From A Form And Insert It Into A Database
// If you spend a lot of time writing pages that take input from a form and insert it into a database, this function will save you time!
// Please Note: You have to name your form fields the same as their corresponding table column is named in mysql for this to work.
// $table - name of the mysql table you are querying
// $exceptions - fields that will not be inserted into table
// i.e. 'submit, action, '; (note trailing comma and space!)
// $sql_type - has to be 'insert' or 'update'
// $sql_condition - have to define this if $sql_type = 'update'
// i.e. "userID = '".$_POST['userID']."'"
function formToDB($table, $exceptions = '', $sql_type = 'insert', $sql_condition = NULL) {
// define some vars
$fields = '';
$values = '';
// format input fields into sql
foreach ($_POST as $field => $value) {
if (!strstr($exceptions,$field)) {
$value = mysql_real_escape_string($value);
if ($sql_type == 'insert') {
$fields .= "$field, ";
$values .= "'$value', ";
}
else {
$fields .= "$field = '$value', ";
}
}
}
// remove trailing ", " from $fields and $values
$fields = preg_replace('/, $/', '', $fields);
$values = preg_replace('/, $/', '', $values);
// create sql statement
if ($sql_type == 'insert') {
$sql = "INSERT INTO $table ($fields) VALUES ($values)";
}
elseif ($sql_type == 'update') {
if (!isset($sql_condition)) {
echo 'ERROR: You must enter a sql condition!';
exit;
}
$sql = "UPDATE $table SET $fields WHERE $sql_condition";
}
else {
echo 'ERROR: Invalid input for argument $sql_type: must be "insert" or "update"';
exit;
}
// execute sql
if (mysql_query($sql)) {
return true;
}
else {
//echo mysql_error();
return false;
}
}





