我如何从datalist获取一个从mysql数据库中获取数据的值? [英] How will i get a value from datalist where data are fetch from mysql database?
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问题描述
所以,我有这个代码可以从数据库中获取数据:
< ?php
// 连接数据库
$ conn = mysqli_connect( localhost, root, , inventory);
if (mysqli_connect_errno($ conn)){
echo 无法连接到数据库服务器;
}
// 查询数据库以填充项目
$ sql = SELECT ITEM_NAME FROM item;
$ query = mysqli_query($ conn,$ sql);
echo ' < datalist name =list1id =list1onchange = showUser(THIS.VALUE)>'跨度>; // 将select更改为datalist
echo ' < option value =>选择项目< / option>';
while ($ selectedItem = mysqli_fetch_assoc($ query)){
echo < option> {$ selectedItem ['ITEM_NAME']}< / option>;
}
echo ' < / datalist>'; // 将select更改为datalist
if (isset($ _POST [' submit'])){
$ selectedItem = $ _POST [' list1'];
}
?>
在此部分中echo< option> {$ selectedItem ['ITEM_NAME']}< / option>;
我正在显示数据库中的数据,但我需要从datalist获得相应的值。
希望听到你们的很多信息!谢谢!
我的尝试:
我试过做就像这样
echo< option value =。$ selectedItem。> {$ selectedItem ['ITEM_NAME']}< / option>;
解决方案
conn = mysqli_connect( localhost, root, , inventory);
if (mysqli_connect_errno(
conn)){
echo 无法连接到数据库服务器;
}
// 查询数据库以填充项目
sql = SELECT ITEM_NAME FROM item;
So, I have this code which can fetch data from database:
<?php
//connect to database
$conn = mysqli_connect("localhost", "root", "", "inventory");
if(mysqli_connect_errno($conn)) {
echo "Unable to connect to database server";
}
//query database for items to populate
$sql = "SELECT ITEM_NAME FROM item";
$query = mysqli_query($conn, $sql);
echo '<datalist name="list1" id="list1" onchange="showUser(this.value)">'; //change select to datalist
echo '<option value="">Select item</option>';
while($selectedItem = mysqli_fetch_assoc($query)) {
echo "<option>{$selectedItem['ITEM_NAME']}</option>";
}
echo '</datalist>'; //change select to datalist
if (isset ($_POST['submit'])) {
$selectedItem = $_POST['list1'];
}
?>
In this part echo "<option>{$selectedItem['ITEM_NAME']}</option>";
I'm displaying the data from the database, but I need to get the corresponding value from the datalist.
Hope to hear a lot from you guys! Thanks!
What I have tried:
I tried to do it like this
echo "<option value=".$selectedItem.">{$selectedItem['ITEM_NAME']}</option>";
解决方案
conn = mysqli_connect("localhost", "root", "", "inventory"); if(mysqli_connect_errno(
conn)) { echo "Unable to connect to database server"; } //query database for items to populate
sql = "SELECT ITEM_NAME FROM item";
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