MYSQL不会对BYDER变量进行排序

Question

I cannot get this Function to ORDER BY a variable. As it appears below, it works perfectly. If I change the word "cabin" to be a variable, even the same variable that already works for this, it doesn't work and returns the error listed below.

Works:

function makedropdown($connection, $table) {
        $result = mysqli_query($connection,"SELECT * FROM $table ORDER BY cabin ASC");
        if (!$result) {
            die ("Database query failed:" . mysqli_error());
        }
        echo "<select name='delcabin'>";
        While ($row=  mysqli_fetch_array($result)){
             echo "<option>" . $row[1] . "</option>";
        }
        echo "</select>";
}

Doesn't work:

$result = mysqli_query($connection,"SELECT * FROM $table ORDER BY $table ASC");

Error Message Received:

mysqli_error() expects exactly 1 parameter, 0 given

Thanks in advance.

Answer 1

Personally, i'd modify the function to accept a third parameter to set the order by variable.

function makedropdown($connection, $table,$OrderBy) {
        $result = mysqli_query($connection,"SELECT * FROM $table ORDER BY $OrderBy ASC");
        if (!$result) {
            die ("Database query failed:" . mysqli_error($connection));
        }
        echo "<select name='delcabin'>";
        While ($row=  mysqli_fetch_array($result)){
             echo "<option>" . $row[1] . "</option>";
        }
        echo "</select>";
}

Then respectively call it by:

$Conn = new mysqli(); // This is a test database connection call, for example only. 
$Tablename = "TestTable"; // This is a valid table within the schema 
$ColumnName = "TestColumn"; // This is a valid column within the SQL Table 

    makedropdown($Conn,$TableName,$ColumnName);

This way, you could avoid your query looking like:

$Table = "Cabin"; 

SELECT * FROM $Table ORDER BY $Table ASC

when echoed out, will print:

SELECT * FROM Cabin ORDER BY Cabin ASC

Answer 2

Just do this...

  $result = mysqli_query($connection,"SELECT * FROM $table ORDER BY '{$yourvariable}' ASC"); 

Also....make sure that your variable is a valid column, otherwise it wont work.

So in this case is $table, being your table name, also a valid column name??