[英]Simple search MySQL database using php
我目前有一個小的 php 腳本,可以根據用戶輸入搜索數據庫。 有一個 html 文件,其中有一個字段用於將搜索字符串輸入到數據庫中。 基本上,您可以搜索員工。
如果找到,它應該檢索員工結果,如果沒有,則應該檢索“未找到員工”消息。
但是,出於某種原因,無論進行何種搜索,查詢都會返回數據庫中的每個員工。
我已經為此工作了一個多小時,老實說我很難過。 這可能是一個簡單的錯誤,但我可以得到一些幫助。
<?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);
?>
任何幫助表示贊賞。
謝謝。
您需要使用$_POST
而不是$_post
。
首先添加HTML代碼:
<form action="" method="post">
<input type="text" name="search">
<input type="submit" name="submit" value="Search">
</form>
現在添加了 PHP 代碼:
<?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"];
}
}
?>
我認為它對每個人都有效
<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>";
}
}
}
?>
只是有了上面的答案,我希望這是問題所在。
$_POST['search']
而不是$_post['search']
再次使用LIKE '%$name%'
而不是LIKE '%{$name}%'
這是一個更好的代碼,可以幫助您完成。
用你的數據庫,而是,我用的是mysql而不是mysqli
好好享受。
<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();
?>
`
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?
如果您執行mysqli_fetch_array()
,則必須將整數放入$row index ex.($row[3])
。如果您讀取$row['id']
或$row['example']
,則必須使用mysqli_fetch_assoc
。
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