Question

我目前有一个小的PHP脚本，可以根据用户输入搜索数据库。有一个html文件，其中有一个字段用于将搜索字符串输入数据库。从本质上讲，您可以搜索员工。

如果找到，则应检索员工结果，如果没有，则应检索“找不到员工”消息。

但出于某种原因，无论搜索如何，查询都会返回数据库中的每个员工。

我一直在研究这个问题超过一个小时，老实说我很难过。这可能是一个简单的错误，但我可以提供一些帮助。

<?php
    $con= new mysqli("localhost","root","","Employee");
    $name = $_post['search'];
    //$query = "SELECT * FROM employees
   // WHERE first_name LIKE '%{$name}%' OR last_name LIKE '%{$name}%'";

    // Check connection
    if (mysqli_connect_errno())
      {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }

$result = mysqli_query($con, "SELECT * FROM employees
    WHERE first_name LIKE '%{$name}%' OR last_name LIKE '%{$name}%'");

while ($row = mysqli_fetch_array($result))
{
        echo $row['first_name'] . " " . $row['last_name'];
        echo "<br>";
}
    mysqli_close($con);
    ?>

任何帮助表示感谢。

感谢。

Answer 1

您需要使用$_POST而不是$_post。

Answer 2

首先添加HTML代码：

<form action="" method="post">
<input type="text" name="search">
<input type="submit" name="submit" value="Search">
</form>

现在添加了PHP代码：

<?php
$search_value=$_POST["search"];
$con=new mysqli($servername,$username,$password,$dbname);
if($con->connect_error){
    echo 'Connection Faild: '.$con->connect_error;
    }else{
        $sql="select * from information where First_Name like '%$search_value%'";

        $res=$con->query($sql);

        while($row=$res->fetch_assoc()){
            echo 'First_name:  '.$row["First_Name"];


            }       

        }
?>

Answer 3

正如上面的回答，我希望这是问题。

$_POST['search']代替$_post['search']

再次使用LIKE '%$name%'代替LIKE '%{$name}%'

Answer 4

这是一个更好的代码，可以帮助您完成。
使用您的数据库，而是我使用mysql而不是mysqli
享受吧。

<body>

<form action="" method="post">

  <input name="search" type="search" autofocus><input type="submit" name="button">

</form>

<table>
  <tr><td><b>First Name</td><td></td><td><b>Last Name</td></tr>

<?php

$con=mysql_connect('localhost', 'root', '');
$db=mysql_select_db('employee');


if(isset($_POST['button'])){    //trigger button click

  $search=$_POST['search'];

  $query=mysql_query("select * from employees where first_name like '%{$search}%' || last_name like '%{$search}%' ");

if (mysql_num_rows($query) > 0) {
  while ($row = mysql_fetch_array($query)) {
    echo "<tr><td>".$row['first_name']."</td><td></td><td>".$row['last_name']."</td></tr>";
  }
}else{
    echo "No employee Found<br><br>";
  }

}else{                          //while not in use of search  returns all the values
  $query=mysql_query("select * from employees");

  while ($row = mysql_fetch_array($query)) {
    echo "<tr><td>".$row['first_name']."</td><td></td><td>".$row['last_name']."</td></tr>";
  }
}

mysql_close();
?>

Answer 5

`

require_once('functions.php');

$errors = FALSE;
$errorMessage = "";

if(mysqli_connect_error()){
  $errors = TRUE;
  $errorMessage .= "There was a connection error <br/>";
  errorDisplay($errorMessage);
  die($errors);
} else if($errors != "TRUE"){
  $errors .= FALSE;
}

if(isset(mysqli_real_escape_string($_POST['search']))){
  $search = mysqli_real_escape_string($_POST['search']);
  search(search);
}



?>
<?php
//This is functions.php
function search($searchQuery){
  echo "<div class="col-md-10 col-md-offset-1">";
  $searchTerm
  $query = query("SELECT * FROM `index` WHERE `keywords` LIKE '".$searchTerm."' ");
  while($row = mysqli_fetch_array($query)){
    $results = <<< DELIMITER
      <div class="result col-md-12">
        <a href="index.php?search={$row['id']}"> {$row['Title']} </a>
        <p class="searchDesc">{$row['description']}</p>
      </div>
DELIMITER;
    echo $results;
  }
  echo "</div>";
}

function errorDisplay($msg){
  if(!isset($_SESSION['errors'])){
    $_SESSION['errors'] = $msg;
    showError($msg);
  } else if() {
    $_SESSION['errors'] .= $msg . "<br>";
    showError($msg);
  }
}
function showError($msg) {
  return $msg;
  unset($_SESSION['errors']);
}


?>`
Perhaps That Helps?

Answer 6

如果执行mysqli_fetch_array()，则必须在$row index ex.($row[3])中放入整数。如果读取$row['id']或$row['example']，则必须使用mysqli_fetch_assoc。

Answer 7

我认为它适合所有人

<html lang="en">

<head>
    <meta charset="UTF-8">
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <title>Search</title>
</head>

<body>
    <form action="" method="post">
        <input type="text" placeholder="Search" name="search">
        <button type="submit" name="submit">Search</button>
    </form>
</body>

</html>
<?php


if (isset($_POST['submit'])) {
    $searchValue = $_POST['search'];
    $con = new mysqli("localhost", "root", "", "testing");
    if ($con->connect_error) {
        echo "connection Failed: " . $con->connect_error;
    } else {
        $sql = "SELECT * FROM customer_info WHERE name OR email LIKE '%$searchValue%'";

        $result = $con->query($sql);
        while ($row = $result->fetch_assoc()) {
            echo $row['name'] . "<br>";
            echo $row['email'] . "<br>";
        }

      
    }   
}



?>

使用php简单搜索MySQL数据库

7 个答案: