Question

我尝试打开这篇文章，以获取有关无法正常工作的搜索表单问题的帮助。

我有一个MySQL DB，其中一些数据存储在不同的表中：

DB是：list_pec 表为：pec_1，pec_2，pec_3和pec_4

所有这些表都包含具有不同数据的相同行。行是 firstame，姓氏，电子邮件，id_client，id_2client

我的目标是在PHP中创建一个搜索表单，其中有一个输入标签和一个选择表单，用于连接到数据库并将其作为输出查询结果。

下面的PHP文件连接我称为“ conn.php”的MySQL DB

<?php
$host = "localhost";
$userName = "demo";
$password = "demo";
$dbName = "list_pec";

// Create database connection
$conn = new mysqli($host, $userName, $password, $dbName);

// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
?>

下面的文件“ search.php”中包含表单和php代码，我想在同一.php文件中获得查询结果，所以我在表单操作中使用了<?pho echo $_SERVER ['PHP_SELF']; ?>

<?
error_reporting(E_ALL);
ini_set('display_errors', '1');
include("conn.php");

$search_output = "";
if (isset($_POST["submit"])){
    if($_POST['option']== "a"){
     $sqlcommand="SELECT email, id_client, id_2client FROM pec_1 WHERE email = 'email'";
     }

     else if ($_POST['option'] == "b"){
     $sqlcommand="SELECT email, id_client, id_2client FROM pec_2 WHERE email = 'email'";
     }

     else if ($_POST['option'] == "c"){
     $sqlcommand="SELECT email, id_client, id_2client FROM pec_3 WHERE email = 'email'";
     }

          else if ($_POST['option'] == "d"){
     $sqlcommand="SELECT email, id_client, id_2client FROM pec_4 WHERE email = 'email'";
     }



$query = mysqli_query($conn,$sqlcommand) or die (mysqli_error($conn));
$search_output .="<hr />query result: ";
if ($row = mysqli_fetch_array($query)){
    $email = $row ["email"];
    $pec = $row ["id_client"];
    $sdi = $row ["id_2client"];
    $search_output .= "<hr/><p> $email - $id_client - $id_2client</p>";


} else{
    $search_output= "<hr /> No Result";

}
}
?>
<html>
<head>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/4.0.0/css/bootstrap.min.css" integrity="sha384-Gn5384xqQ1aoWXA+058RXPxPg6fy4IWvTNh0E263XmFcJlSAwiGgFAW/dAiS6JXm" crossorigin="anonymous">

<title>Search id_client and id_2client</title>
</head>
<body>
    <section>
        <div class="container">
            <div class="row my-5">
                <h1>Search id_client and id_2client</h1>
            </div>
<form name="ricerca-pec" method="post" action="<?php echo $_SERVER ['PHP_SELF']; ?>">
  <div class="form-group">
    <label for="exampleFormControlInput1">Inserisci l'email del cliente</label>
    <input type="email" class="form-control" id="exampleFormControlInput1" placeholder="youremail@email.com" name="email">
  </div>
  <div class="form-group">
    <label for="exampleFormControlSelect1">Select Option</label>
    <select class="form-control" id="exampleFormControlSelect1" name="option">
      <option value="a">A</option>
      <option value="b">B</option>
      <option value="c">C</option>
      <option value="d">D</option>
    </select>
  </div>
  <input class="btn btn-primary" type="submit" name="submit"></input>
</form>

</div>
</section>

<section>
<div class="container">
<div class="row">
<p><?php echo $search_output; ?></p>
</div>
</div>
</section>


<script src="https://code.jquery.com/jquery-3.2.1.slim.min.js" integrity="sha384-KJ3o2DKtIkvYIK3UENzmM7KCkRr/rE9/Qpg6aAZGJwFDMVNA/GpGFF93hXpG5KkN" crossorigin="anonymous"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/popper.js/1.12.9/umd/popper.min.js" integrity="sha384-ApNbgh9B+Y1QKtv3Rn7W3mgPxhU9K/ScQsAP7hUibX39j7fakFPskvXusvfa0b4Q" crossorigin="anonymous"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/4.0.0/js/bootstrap.min.js" integrity="sha384-JZR6Spejh4U02d8jOt6vLEHfe/JQGiRRSQQx"</script>
</body>
</html>

当我使用搜索表单（在文件/search.php上）时，我得到一个“无结果”的答案，因此我看到查询已正确执行，但是似乎变量'email'不是从$发送的_POST在表单上提交以进行查询。

实际上，如果我在search.php文件中进行修改，则第一个查询替换为“ email”，并替换为“ email@email.com”（包含在表DB pec_1中），我可以从查询中看到正确的结果< / p>

     $sqlcommand="SELECT email, id_client, id_2client FROM pec_1 WHERE email = 'email@email.com'";

请，我要求一些帮助来理解问题并解决，我在没有解决问题的情况下阅读了其他文章。

谢谢。

Answer 1

如果您无法提取任何电子邮件，那是因为您在r中以明文方式传递了该电子邮件。例如：

$sqlCommand

应该更像：

$sqlcommand="SELECT email, id_client, id_2client FROM pec_4 WHERE email = 'email'";

但是您尚未定义任何电子邮件变量，因此它将不起作用。您的代码非常混乱，并且存在一些安全问题。我给您一些提示，作为您的PHP部分的注释，以使它更好地工作。

$sqlcommand="SELECT email, id_client, id_2client FROM pec_4 WHERE email = {$email}";

还要回答有关表单的问题，当脚本位于同一页面中时，简单的<?php // Require only once the connection to avoid multiples connexions require_once('conn.php'); // Debug here error_reporting(E_ALL); ini_set('display_errors', 1); $search_output = ""; // Check if the form has been sent if (isset($_POST["submit"])){ // Define variables $table = ''; $option = ''; $email = ''; // Check if an option is selected if(!isset($_POST["option"])) { // ToDo: Security - check user input $option = $_POST["option"]; } else { // Throw error to the user } // Check if a email is set if(!isset($_POST["email"])) { // ToDo: Security - check user input $email = $_POST["email"]; } else { // Throw error to the user } // Use a switch instead of plenty if switch ($option) { case "a": $table = "pec_1"; break; case "b": $table = "pec_2"; break; case "c": $table = "pec_3"; break; case "d": $table = "pec_4"; break; default: $table = "pec_1"; break; } // write your query once $selectQuery = "SELECT email, id_client, id_2client FROM {$table} WHERE email LIKE %{$email}%"; // Perform your query $query = mysqli_query($conn, $selectQuery); // Fetch the result as array $result = mysqli_fetch_array($query); $search_output .="<hr />query result: "; // If no results if(count($results) < 1) { $search_output= "<hr /> No Result"; } else { $email = $result["email"]; $pec = $result["id_client"]; $sdi = $result["id_2client"]; $search_output .= "<hr/><p> {$email} - {$pec} - {$sdi}</p>"; } } ?>属性足以重新加载同一页面。

PHP表格，用于在MySQL数据库中搜索数据

1 个答案: