Php 按钮提交,从数据库中获取其他按钮
这是我目前的代码:Php 按钮提交,从数据库中获取其他按钮,php,Php,这是我目前的代码: require_once 'functions.php'; // makes the connection to the server to get the state button names $query = "SELECT state FROM state"; $result = $connection->query($query); if ($result === false) { // error mssg echo "<p>
require_once 'functions.php';
// makes the connection to the server to get the state button names
$query = "SELECT state FROM state";
$result = $connection->query($query);
if ($result === false) {
// error mssg
echo "<p>Query fout.</p>";
}
// button of the state gets the buttons of the city
if (isset($_POST['state'])) $state = $_POST['state']; {
query = "SELECT city FROM city='$cityid'";
$result = $connection->query($query);
} if ($result === false) {
// geef nette foutmelding
echo "<p>Query fout.</p>";
}
<?php
//gives the result to Submit html
while ($row = $result->fetch_assoc()) {
echo "<input type ='submit' name='provincie' value='".$row['provincie']."'>";
}
$result->free();
?>
</for
require_once'functions.php';
//与服务器建立连接以获取状态按钮名称
$query=“从状态中选择状态”;
$result=$connection->query($query);
如果($result==false){
//误差mssg
回声“查询fout.”;
}
//国家的按钮得到城市的按钮
如果(isset($_POST['state']))$state=$_POST['state'];{
query=“从城市中选择城市='$cityid';
$result=$connection->query($query);
}如果($result==false){
//geef nette熔接
回声“查询fout.”;
}
您需要单独执行查询结果,因此应该有两个while语句
这不是最好的办法,但我有一个办法:
require_once 'functions.php';
$query = "SELECT state FROM state";
if($query->num_rows > 0){
while($row= $result->fetch_assoc()) {
echo "<input type ='submit' name='provincie' value='".$row['provincie']."' onclick="sendVal(this.value)" >";
}
}
<script type="text/javascript">
function sendVal(item) {
document.cookie="city=" + item + ";";
location.reload();
}
</script>
<?php
if(!is_null($_COOKIE['city'])){
$retrievedcities= "SELECT city FROM city WHERE city ='".$_COOKIE['city']."'";
$con->query($retrievedcities);
while ($row = $result->fetch_assoc()) {
echo "<input type ='submit' name='citybut' value='".$row['city']."'>";
}
unset($_COOKIE['city']);
}
require_once'functions.php';
$query=“从状态中选择状态”;
如果($query->num\u rows>0){
而($row=$result->fetch_assoc()){
回声“;
}
}
函数sendVal(项){
document.cookie=“city=“+item+”;”;
location.reload();
}
这个代码部分工作吗?这似乎应该是错误的if(isset($\u POST['state']))$state=$\u POST['state'];{
。我想你应该用AJAX来做的事情。发送一个请求,获取响应,根据初始返回的数据发送另一个请求,对吗?是的,它不需要:“//状态按钮获取…$result=$connection->query($query);”。我把代码放在上面,以显示我已经尝试过的内容。我尝试了很多谷歌,但我不知道该问谷歌什么来解决我自己的问题。因为我不只是在寻找一个脚本来复制和过去。