我的 PHP 从数据库中的 table 搜索名字或姓氏的代码没有在我的网页上打印任何内容

Question

用户假设在网页的搜索栏中输入名字或姓氏，并假设将 table 中的属性列出到网页中。无论我在搜索栏中输入什么，都不会输出任何内容。我观看了 this 有关如何在 php 中搜索的视频。我试着看了一个多小时，但我找不到任何错误。我的网页中没有收到任何错误消息。

<?php
$serverName = 'localhost';
$userName = 'root';
$password = '';
$databaseName = 'project3';

$connection = mysqli_connect($serverName, $userName, $password, 
$databaseName);
if (!$connection) {
  die("Connection Failed: " . mysqli_connect_error());
  }
echo "Connected Successfully!! <br>";
$output = '';
if (isset($_Post['search'])) {
  $searchq = $_Post['search'];
  $searchq = preg_replace("#[^0-9a-z]#i", "", $searchq);

  $query = mysqli_query("SELECT * from employee WHERE fname LIKE 
  '%$searchq%' OR"
        . "lname LIKE '%$searchq%") or die("failed");
$count = mysqli_num_rows($query);
if ($count == 0) {
    $output = 'No search results';
} else {
    while ($row = mysqli_fetch_array($query)) {
        $firstname = $row['fname'];
        $lastname = $row['lname'];
        $id = $row['id'];
        $output .= '<div>' . $firstname . '' . $lname . '</div>';
        echo "hi";
    }
 }
}

?>

<!DOCTYPE html>

 <html>
<head>
    <meta charset="UTF-8">
    <title>Database Webpage</title>
<font color ="white">
<h1 style="background-color:black; text-align: center">Datebase Website</h1>

 <font color ="black">
</head>
 <body>

   <form action = "index.php" method = "POST">
      <input type = "text" name ="search" placeholder="Search"/>
      <input type= "submit" value = ">>"/>


     </form>

    <?php print("$output"); ?>
 </body>
 </html>

Answer 1

mysqli_query 确实需要 2 个参数，因为 the PHP api 显示了您拥有的程序形式。

因此，对于那部分，调用应如下所示：

$query = mysqli_query($connection, "SELECT * from employee WHERE fname LIKE 
           '%$searchq%' OR"
    . "lname LIKE '%$searchq%") or die("failed");

但是查询会有问题。请注意查询的这一部分：

           '%$searchq%' OR" <--- No space after the OR
    . "lname LIKE '%$searchq%")
       ^--- No space before lname

这两个区域中的任何一个都需要space。

查询字符串的最后一部分最终将如下所示：

'%$searchq%' ORlname LIKE '%$searchq%
^--- trouble (no space) ^---and trouble here (missing a closing single quote)

有时单独设置查询很有用，这样您就可以回显它以检查它在关键字、列、值、逗号用法（需要时）、正确引用等方面的句法是否正确，等等

考虑这个区别：

$query = "SELECT * from employee WHERE fname LIKE '%$searchq%' OR "
       . "lname LIKE '%$searchq%'";
// query check (delete after validation, or comment-out)
echo $query;

$result = mysqli_query($connection, $query);
// I like to use $result, as the query call will return a result set with SELECT
// or false with failure. (though it can return true for other query types, see the api link)

将错误作为死消息的一部分输出也很有帮助：

if (!$result) { // Doh! something wrong...
  die('failed: ' . mysqli_error($connection));
} else {
  $count = mysqli_num_rows($result); // check the result count
  if ($count == 0) {
    $output = 'No search results';
  } else {
    while ($row = mysqli_fetch_array($result)) { // fetch a row
        $firstname = $row['fname'];
        $lastname = $row['lname'];
        $id = $row['id'];
        $output .= '<div>' . $firstname . '' . $lname . '</div>';
        echo "hi";
    }
  }
}

HTH

我的 PHP 从数据库中的 table 搜索名字或姓氏的代码没有在我的网页上打印任何内容

My PHP code to search for a first name or last name from a table in a database is not printing anything to my webpage

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