如何从SQL表中的列中获取所有数据，并将其与从表单中获取的数据进行核对？到目前为止，我拥有的是：
<?php
$con=mysqli_connect("www.tqbtest.comlu.com","a5349216","Password","a5349216_test");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
//How do I check the data from the form to make sure there is no username/email already used?
$sql="INSERT INTO Profiles (firstName, lastName, username, email, password, region, profileGroup, verified)
VALUES
('$_POST[firstname]','$_POST[lastname]','$_POST[age]')";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo "1 record added";
mysqli_close($con);
?>
您有两个选项：
- 您可以将唯一索引添加到任何希望唯一的字段。它们将导致任何违反唯一性约束的插入失败，然后您可以捕获这些插入并将其反馈给应用程序
- 您可以尝试在表中选择希望唯一的字段，看看是否返回结果。如果您这样做，您可以向用户返回错误
我认为这就是您所需要的：
$DBsql = mysql_query("select col1, col2 from table_name where some_id = $some_id");
$sql = mysql_fetch_array($DBsql);
if(($_POST['field1'] == $sql['col1']) && ($_POST['field2'] == $sql['col2'])) {
echo 'The fields are valid';
} else {
echo 'One or more fields are not valid';
}