在mysql数据库中搜索数据的php表单

我试着打开这篇文章来获得一些关于搜索表单出现问题的帮助。
我有一个mysql数据库，其中一些数据存储在不同的表中：
db is:列表\pec表格为：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。行为firstname、lastname、email、id\u client、id\u 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作为输出查询结果提供给我。
下面的php文件连接mysql数据库，我称之为“conn.php”

<?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);
}
?>

下面是一个名为“search.php”的文件，里面有表单和php代码，obv我希望查询结果与我使用的.php文件相同 <?pho echo $_SERVER ['PHP_SELF']; ?> 形式行动

<?
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中）时，我得到了一个答案“no result”，因此我看到查询执行正确，但变量“email”似乎不是从表单submit to query上的$\u post发送的。
实际上，如果我在search.php文件中修改第一个查询，将其替换为“email”和email@email.com'（包含在表dbpec1中）我可以从查询中看到正确的结果

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

请帮我理解问题并解决问题，我看了其他帖子却没有解决我的问题。
谢谢您。

如果你拿不到任何电子邮件，那是因为你在你的邮箱里直接传递了它 $sqlCommand . 例如：

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

应该更像：

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

但是您还没有定义任何电子邮件变量，所以它不会工作。你的代码很混乱，有一些安全问题。我给了您一些提示，作为对php部分的注解，以使其更好地工作。

<?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>";
    }

}
?>

也要回答关于窗体的问题，当脚本位于同一页时 action="/" 属性足以重新加载同一页。

在mysql数据库中搜索数据的php表单

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