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php - 来自数据库同一表的 PHP 中的相关下拉列表

转载 作者:行者123 更新时间:2023-11-29 02:46:04 25 4
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我需要制作一些依赖下拉列表。在这里,从第一个下拉列表中,每当我选择 Region 时,在第二个下拉列表的选项中,它应该显示所选 Region 的相应 Country。但是在选择一个区域 之后,第二个下拉列表没有显示任何内容。我搜索了解决方案,但没有找到合适的解决方案,因为我只针对一个数据库表执行所有这些操作。

数据库

SET SQL_MODE = "NO_AUTO_VALUE_ON_ZERO";
SET time_zone = "+00:00";

CREATE TABLE `geo` (
`Region` varchar(200) NOT NULL,
`Country` varchar(200) NOT NULL,
`City` varchar(200) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

INSERT INTO `geo` (`Region`, `Country`, `City`) VALUES
('Asia', 'Bangladesh', 'Dhaka'),
('Europe', 'France', 'Paris'),
('Asia', 'Bangladesh', 'Khulna'),
('Europe', 'France', 'Avignon'),
('Europe', 'Spain', 'Barcelona'),
('Europe', 'Spain', 'Madrid'),
('Asia', 'Srilanka', 'Colombo');

dbconnect,php

<?php
error_reporting( ~E_DEPRECATED & ~E_NOTICE );

define('DBHOST', 'localhost');
define('DBUSER', 'root');
define('DBPASS', '');
define('DBNAME', 'ddl');

$conn = mysql_connect(DBHOST,DBUSER,DBPASS);
$dbcon = mysql_select_db(DBNAME);

if ( !$conn ) {
die("Connection failed : " . mysql_error());
}

if ( !$dbcon ) {
die("Database Connection failed : " . mysql_error());
}

?>

index.php

<?php
require_once 'dbconnect.php';

$query ="SELECT * FROM geo";
$results = mysql_query($query);
?>
<!DOCTYPE html>
<html>
<head>
<title>DDL - Select</title>
</head><?php
$bul[10000007] = false;
?>

<style>
body{width:610px;}
.frmDronpDown {border: 1px solid #F0F0F0;background-color:#C8EEFD;margin: 2px 0px;padding:40px;}
.demoInputBox {padding: 10px;border: #F0F0F0 1px solid;border-radius: 4px;background-color: #FFF;width: 50%;}
.row{padding-bottom:15px;}
</style>
<script src="https://code.jquery.com/jquery-2.1.1.min.js" type="text/javascript"></script>
<script>
function getCountry(val) {
$.ajax({
type: "POST",
url: "get_country.php",
data:'region_id='+val,

success: function(data){
$("#country-list").html(data);
}
});
}

function selectRegion(val) {
$("#search-box").val(val);
$("#suggesstion-box").hide();
}
</script>

<body>
<div class="frmDronpDown">
<div class="row">

<form>
<label>Region: </label>
<select name="region" id="region-list" class="demoInputBox" onChange="getCountry(this.value);">
<option value="">Select Region</option>

<?php

$sql = "SELECT * FROM geo";
$res = mysql_query($sql);
while($row = mysql_fetch_assoc($res)) {

if($bul[$row['Region']] != true && $row['Region'] != 'Region'){
?>
<option value="<?php echo $row['Region']; ?>"><?php echo $row['Region']; ?></option>
<?php
$bul[$row['Region']] = true;
}
}
?>

</select>
</div>


<div class="row">
<label>Country: </label>
<select name="country" id="country-list" class="demoInputBox">
<option value="">Select Country</option>
</select>


</form>

</div>
</div>

</body>
</html>

get_country.php

<?php
require_once 'dbconnect.php';


if(!empty($_REQUEST["region_id"])) {

$query ="SELECT * FROM geo WHERE Region =" . $_POST["region_id"];
$result = mysql_query($query);
?>
<option value="">Select Country</option>

<?php
while($row2=mysql_fetch_assoc($result)){
if($bul2[$row2['Country']] != true && $row2['Country'] != 'Country') { ?>
<option value="<?php echo $row['Country']; ?>"><?php echo $row['Country']; ?> </option>
<?php
$bul2[$row2['Country']] = true;
}
}
}
?>

最佳答案

使用 mysqli 函数而不是 mysql 函数。并且还使用 mysqli_escape_string() 函数来防止 sql 注入(inject)。

在 get_country.php 文件中你错过了 $_POST["region_id"] 前后的 '

$query ="SELECT * FROM geo WHERE Region =" . "'" . mysqli_escape_string($conn, $_POST["region_id"] ) ."'";

而且您还需要在 $row['Country'] 中修复,它将是 $row2['Country']。

这是我机器上的工作代码:

索引.php

<?php
require_once 'dbconnect.php';
$query = "SELECT * FROM geo";
$results = mysqli_query($conn, $query);
?>
<!DOCTYPE html>
<html>
<head>
<title>DDL - Select</title>
</head><?php
$bul[10000007] = false;
?>

<style>
body{width:610px;}
.frmDronpDown {border: 1px solid #F0F0F0;background-color:#C8EEFD;margin: 2px 0px;padding:40px;}
.demoInputBox {padding: 10px;border: #F0F0F0 1px solid;border-radius: 4px;background-color: #FFF;width: 50%;}
.row{padding-bottom:15px;}
</style>
<script src="https://code.jquery.com/jquery-2.1.1.min.js" type="text/javascript"></script>
<script>
function getCountry(val) {
$.ajax({
type: "POST",
url: "get_country.php",
data:'region_id='+val,

success: function(data){
$("#country-list").html(data);
}
});
}

function selectRegion(val) {
$("#search-box").val(val);
$("#suggesstion-box").hide();
}
</script>

<body>
<div class="frmDronpDown">
<div class="row">

<form>
<label>Region: </label>
<select name="region" id="region-list" class="demoInputBox" onChange="getCountry(this.value);">
<option value="">Select Region</option>

<?php
$sql = "SELECT * FROM geo";
$res = mysqli_query($conn, $sql);
while ($row = mysqli_fetch_assoc($res)) {
if ($bul[$row['Region']] != true && $row['Region'] != 'Region') {
?>
<option value="<?php
echo $row['Region'];
?>"><?php
echo $row['Region'];
?></option>
<?php
$bul[$row['Region']] = true;
}
}
?>

</select>
</div>

<div class="row">
<label>Country: </label>
<select name="country" id="country-list" class="demoInputBox">
<option value="">Select Country</option>
</select>

</form>
</div>
</div>
</body>
</html>

get_country.php

<?php
require_once 'dbconnect.php';


if(!empty($_REQUEST["region_id"])) {

$query ="SELECT * FROM geo WHERE Region =" . "'" . mysqli_escape_string($conn, $_POST["region_id"] ) ."'";
//echo $query ="SELECT * FROM geo WHERE Region =" . "'" . $_POST["region_id"] ."'";

$result = mysqli_query($conn, $query);
?>
<option value="">Select Country</option>

<?php
while($row2=mysqli_fetch_assoc($result)){

//var_dump($row2);
if($bul2[$row2['Country']] != true && $row2['Country'] != 'Country' || 1) { ?>
<option value="<?php echo $row2['Country']; ?>"><?php echo $row2['Country']; ?> </option>
<?php
$bul2[$row2['Country']] = true;
}
}
}
?>

和 dbconnect.php

<?php
error_reporting( ~E_DEPRECATED & ~E_NOTICE );

define('DBHOST', 'localhost');
define('DBUSER', 'root');
define('DBPASS', '');
define('DBNAME', 'ddl');

$conn = mysqli_connect(DBHOST,DBUSER,DBPASS);
$dbcon = mysqli_select_db($conn, DBNAME);

if ( !$conn ) {
die("Connection failed : " . mysqli_error());
}

if ( !$dbcon ) {
die("Database Connection failed : " . mysqli_error());
}

?>

关于php - 来自数据库同一表的 PHP 中的相关下拉列表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42088706/

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