Question

修正：感谢大家的帮助，对于那些需要从某个地方开始的人，你可以将它作为参考

我正在尝试学习PHP SQL HTML，我正在努力正确运行我的代码，你能帮我解决吗？我已经看了很多例子，但是我不能让它工作，我的html表格没有显示出来。这是一个程序，它将搜索用户在搜索表单中输入的内容，并在sql数据库中查找。

     <html>
  <head>
    <meta  http-equiv="Content-Type" content="text/html;  charset=iso-8859-1">
    <title>Search  Contacts</title>
  </head>
  <body>
    <h3>Search  Contacts Details</h3>
    <p>You  may search either by first or last name</p>
    <form  method="post" action="#"  id="searchform">
      <input  type="text" name="name">
      <input  type="submit" name="submit" value="Search">
    </form>

<?php
$servername = "xxx";
$username = "xxx";
$password = "xxx";
$dbname = "Staff";
$myquery = $_POST["name"];

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);

// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());

} // SQL查询

if (!empty($_POST["name"])) {
    $sql = "select ID, FirstName, LastName, Email, PhoneNumber from Userlist where FirstName LIKE '%". $myquery  ."%'" ;

    //Get query on the database
    $result = mysqli_query($conn, $sql);

    //Check results
    if (mysqli_num_rows($result) > 0)
    {
      while($row = mysqli_fetch_assoc($result))
        {
          echo "id:" . $row["ID"] . 
               " - Name: " . $row["LastName"]. " " . $row["FirstName"]. // Correct here 
               " - Email: " . $row["Email"] . 
               " - PhoneNumber: " . $row["PhoneNumber"]. "<br>";
        }
    } else {
        echo "0 results";
    }
}

    mysqli_close($conn);

?>
</body>
</html>

Answer 1

您需要使用'到LIKE更正SQL语句。

$sql = "select ID, FirstName, LastName, Email, PhoneNumber  
       from Userlist where FirstName LIKE '%". $myquery  ."%'" ;

另外，当您在$myquery上$_POST变量时，我认为您应该执行以下操作：

if (!empty($_POST["name"])) {

    $myquery = $_POST["name"];

    $sql = "select ID, FirstName, LastName, Email, PhoneNumber 
           from Userlist where FirstName LIKE '%". $myquery  ."%'" ;

    //Get query on the database
    $result = mysqli_query($conn, $sql);

    //Check results
    if (mysqli_num_rows($result) > 0)
    {
      while($row = mysqli_fetch_assoc($result))
        {
          echo "id:" . $row["ID"] . 
               " - Name: " . $row["LastName"]. " " . $row["FirstName"]. // Correct here 
               " - Email: " . $row["Email"] . 
               " - PhoneNumber: " . $row["PhoneNumber"]. "<br>";
        }
    } else {
        echo "0 results";
    }
}

并确保 您的表单操作与您的PHP代码所属的当前页面相同。

Answer 2

在if（！$ conn）{}

之后缺少结束大括号

Answer 3

尝试使用每行输出的回声。其次要注意“建议”。第三，为了安全起见，转义你的post变量。

简单的搜索程序PHP / SQL

3 个答案: