How list out all tables in MSSQL?

I'm using the code below to show the tables in my database.

I get "Connected to database" but nothing else. Is my code correct? can I use another way to get the info I need?

<?php 
$link = m开发者_开发问答ssql_connect('HOST', 'user', 'pass');

if (!$link || !mssql_select_db('dbname', $link)) {
    die('Unable to connect or select database!');
}else{
echo"Connected to database";
}


$v = mssql_query("Select name from sysobjects where type like 'u'");
$row = mssql_fetch_array($v);

echo "<br>";  echo $row[0]; echo "<br>";


mssql_free_result($v);
?>

Alternate way, also fetches schema name

SELECT TABLE_CATALOG ,
        TABLE_SCHEMA ,
        TABLE_NAME ,
        TABLE_TYPE
FROM INFORMATION_SCHEMA.TABLES

SELECT * FROM sys.Tables;

Should do the magic :-D

And if u want to see all columns, i would do

SELECT TOP 1 * From Tablename;

so u'll get one row with all Columns, its not perfect but it does the trick if u just want to know sth

I do this:

// Check if table exists
        $sqlExist = "SELECT * FROM INFORMATION_SCHEMA.TABLES WHERE TABLE_NAME = 'dem';";
        $stmtExist = sqlsrv_query( $conn_mssql, $sqlExist );
        $exist = $row = sqlsrv_fetch_array( $stmtExist, SQLSRV_FETCH_ASSOC);

        if ($exist == "") {
          echo "Die abgefragte Tabelle existiert nicht oder ist nicht erreichbar!";
        } else {
          // MSSQL QUERY ITSELF
          $sql = "SELECT * from dem;";
          $stmt = sqlsrv_query( $conn_mssql, $sql );
          if( $stmt === false) {
              die( print_r( sqlsrv_errors(), true) );
          }

          while( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_ASSOC) ) {
            echo $row['id'] . " " . $row['cdemo2'] . ", " . $row['cdemo1'] . "<br>";
          }

          sqlsrv_free_stmt( $stmt);

        }