MySQLi Procedural create Table frage

[Loddar1]

Wenn ich meine Tabellen installieren will, geht das bisher so:

<?php
$servername = "xxxxxx";
$username = "xxxxx";
$password = "xxxxxx";
$dbname = "xxxxxx";

$sql_link = mysqli_connect($servername, $username, $password, $dbname);
if (!$sql_link) {
die("Connection failed: " . mysqli_connect_error());
}

$sqlquerya = "CREATE TABLE `sites` (
  `mid` int(4) unsigned NOT NULL auto_increment,
  `name` varchar(20) COLLATE utf8_unicode_ci NOT NULL,
  `heading` varchar(30) COLLATE utf8_unicode_ci NOT NULL,
  PRIMARY KEY  (`mid`)) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;";

$sqlqueryb = "CREATE TABLE `meta` (
     `id` int(3) NOT NULL default '1',
     `descript` varchar(200) COLLATE utf8_unicode_ci NOT NULL,
     `keyword` varchar(200) COLLATE utf8_unicode_ci NOT NULL,
     PRIMARY KEY  (`id`)) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;";
   
$sqlqueryc = "CREATE TABLE `impres` (
  `id` int(11) NOT NULL default '1',
  `vname` varchar(25) COLLATE utf8_unicode_ci NOT NULL,
  `nname` varchar(30) COLLATE utf8_unicode_ci NOT NULL,
  PRIMARY KEY (`id`)) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;";

$sqlqueryd = "CREATE TABLE `news` (
  `id` int(10) unsigned NOT NULL auto_increment,
  `titel` varchar(250) COLLATE utf8_unicode_ci NOT NULL,
  `name` varchar(250) COLLATE utf8_unicode_ci NOT NULL,
  PRIMARY KEY (`id`)) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;";



if (mysqli_query($sql_link, $sqlquerya)) {
        echo "Table sites created successfully<br/>";
    }
    else
    {
    echo "Error creating table: " . mysqli_error($sql_link);
}

if (mysqli_query($sql_link, $sqlqueryb)) {
        echo "Table meta created successfully<br/>";
        }
        else
        {
    echo "Error creating table: " . mysqli_error($sql_link);
}

if (mysqli_query($sql_link, $sqlqueryc)) {
            echo "Table impres created successfully<br/>";
        }
        else
        {
    echo "Error creating table: " . mysqli_error($sql_link);
}
if (mysqli_query($sql_link, $sqlqueryd)) {
            echo "Table news created successfully<br/>";
        }
        else
        {
    echo "Error creating table: " . mysqli_error($sql_link);
}

mysqli_close($sql_link);
?>

Jetzt wollte ich es verkürzen und übersichtlicher machen,
aber da bekomme ich Fehler gemeldet:
irgendwas mit "default '1'" darf nicht sein.

Wenn ich meine Tabellen füllen will, dann geht die 2te option mit $ss .= xxx

<?php
$servername = "xxxxxx";
$username = "xxxxx";
$password = "xxxxxx";
$dbname = "xxxxxx";

$sql_link = mysqli_connect($servername, $username, $password, $dbname);
if (!$sql_link) {
die("Connection failed: " . mysqli_connect_error());
}

$sqlquery = "CREATE TABLE `sites` (
  `mid` int(4) unsigned NOT NULL auto_increment,
  `name` varchar(20) COLLATE utf8_unicode_ci NOT NULL,
  `heading` varchar(30) COLLATE utf8_unicode_ci NOT NULL,
  PRIMARY KEY  (`mid`)) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;";

$sqlquery .= "CREATE TABLE `meta` (
     `id` int(3) NOT NULL default '1',
     `descript` varchar(200) COLLATE utf8_unicode_ci NOT NULL,
     `keyword` varchar(200) COLLATE utf8_unicode_ci NOT NULL,
     PRIMARY KEY  (`id`)) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;";
   
$sqlquery .= "CREATE TABLE `impres` (
  `id` int(11) NOT NULL default '1',
  `vname` varchar(25) COLLATE utf8_unicode_ci NOT NULL,
  `nname` varchar(30) COLLATE utf8_unicode_ci NOT NULL,
  PRIMARY KEY (`id`)) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;";

$sqlquery .= "CREATE TABLE `news` (
  `id` int(10) unsigned NOT NULL auto_increment,
  `titel` varchar(250) COLLATE utf8_unicode_ci NOT NULL,
  `name` varchar(250) COLLATE utf8_unicode_ci NOT NULL,
  PRIMARY KEY (`id`)) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;";


if (mysqli_query($sql_link, $sqlquery)) {
        echo "Tables created successfully<br/>";
    }
    else
    {
    echo "Error creating tables: " . mysqli_error($sql_link);
}

mysqli_close($sql_link);
?>

Sieht einer einen Fehler oder kann mir helfen?
Danke schonmal,
Die Tabellen sind verkürzt dargestellt.

 

[Tommy57]

Ich würde jetzt sagen, vllt weil du versucht einen String als Default Wert auf ein Integer Feld zu schreiben?!
Probier doch mal aus, die Hochkomma weg zu lassen?