Question

我仍然不熟悉从带有php的html网页表格到数据库的POSTING，但我想我已经能够拼凑出我需要的一切。但是仍然缺少某些东西，我不知道是什么。

我测试了我的连接和表格：

<?php
  $link = mysqli_connect("localhost","#user","#password", "#database");
  if (!$link) {
    echo "Error: Unable to connect to MySQL." . PHP_EOL;
    echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
    echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
    exit;
  }
  echo "Success: A proper connection to MySQL was made! The my_db database is great." . PHP_EOL;
  echo "Host information: " . mysqli_get_host_info($link) . PHP_EOL;
  $test_query = "SHOW TABLES FROM" $#database;
  $result = mysqli_query($connect, $test_query);
  $tblCnt = 0;
  while($tbl = mysqli_fetch_array($result)) {
    $tblCnt++;
    #echo $tbl[0]."<br />\n";}
    if (!$tblCnt) {
        echo "There are no tables<br />\n";
    } else {
        echo "There are $tblCnt tables<br />\n";
    }
  mysqli_close($link);
?>

我尝试过127.0.0.1并且访问主机IP（RemoteMySQL）无济于事。我的用户拥有完全权限。我不知道任何其他可能妨碍我的默认安全性。

我使用MySQL和phpmyadmin托管Bluehost。我的数据库中有2个表，当我在phpmyadmin中输入“SHOW TABLES”时会找到它们。我错过了什么？这是我所不知道的。我的智慧结束了......

非常感谢你！

Answer 1

我在这一行中看到了一些问题。

$test_query = "SHOW TABLES FROM" $#database; // Line 10

您没有任何名为$ database的变量。并且您的符号＃不正确。＃用于评论（http://php.net/manual/en/language.basic-syntax.comments.php）

FROM之后缺少空间。这一行应该是

$test_query = "SHOW TABLES FROM " . $database;

否则，生成的查询将是“SHOW TABLES FROMDATABASE_NAME”，这将触发SQL错误。

此外，在while循环后缺少大括号。这是完整的代码更正。

<?php
$link = mysqli_connect("localhost","#user","#password", "#database");
if (!$link) {
    echo "Error: Unable to connect to MySQL." . PHP_EOL;
    echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
    echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
    exit;
 }
 echo "Success: A proper connection to MySQL was made! The my_db 
 database is great." . PHP_EOL;
 echo "Host information: " . mysqli_get_host_info($link) . PHP_EOL;
 $database = "YOUR_DATABASE_NAME_HERE";
 $test_query = "SHOW TABLES FROM " . $database;
 $result = mysqli_query($link, $test_query);
 $tblCnt = 0;
 while($tbl = mysqli_fetch_array($result)) {
     $tblCnt++;
 }
 // echo $tbl[0]."<br />\n";}
 if (!$tblCnt) {
     echo "There are no tables<br />\n";
 } else {
     echo "There are $tblCnt tables<br />\n";
 }
 mysqli_close($link);
 ?>

我希望它有所帮助。

表显示在phpmyadmin但不查询+更多

1 个答案: