Javascript 从数据库检索数据并显示在文本框中
如何将数据库中的数据显示到文本框中?请帮忙Javascript 从数据库检索数据并显示在文本框中,javascript,php,mysql,Javascript,Php,Mysql,如何将数据库中的数据显示到文本框中?请帮忙 //Javascript textbox <div class="Text"> <input class="Text" type="text" value=" <?PHP echo $id?>" name="id" size="19"/> //PHP MYSQL Connect c
//Javascript textbox
<div class="Text">
<input class="Text" type="text" value="
<?PHP echo $id?>" name="id" size="19"/>
//PHP MYSQL Connect code
<?php
error_reporting(0);
include('../connection.php');
$id =$_REQUEST['id'];
$result = mysql_query("SELECT * FROM cust WHERE id = '$id'");
$test = mysql_fetch_array($result);
if (!$result)
{
die("Error: Data not found..");
}
$id=$test['id'] ;
?>
//Javascript文本框
将PHP代码放在HTML之前
//PHP MYSQL Connect code
<?php
error_reporting(0);
include('../connection.php');
$id =$_REQUEST['id'];
$result = mysql_query("SELECT * FROM cust WHERE id = '$id'");
$test = mysql_fetch_array($result);
if (!$result)
{
die("Error: Data not found..");
}
$id=$test['id'] ;
?>
//Javascript textbox
<div class="Text">
<input class="Text" type="text" value="
<?PHP echo $id?>" name="id" size="19"/>
//PHP MYSQL连接代码
将以下PHP代码放在HTML之前
PHP
HTML和PHP(正文内部)
元素呈现的一个好方法是创建HTML辅助程序库。例如,创建一个包含一组静态标记创建者方法的类HTML
#Pseudo HTML helper class - HTML.class.php
class HTML
public static input(type, id, class, data, text)
public static heading(mode, text)
#Pseudo input tag helper - HTML.class.php::input
function input(type, id, value) {
# method create the html string for the given input.
return ["<input type=",type," id=",id," value=",value,"/>"].join('');
}
<?php
$con = new mysqli(host,user,pass,dbname);
$query = "SELECT * FROM cust WHERE id = '$id'";
$result = $con-> query($query);
while ($row = $result->fetch_assoc()){
$value = $row['id'];
echo HTML::input('text', id, $id);
}
?>
#伪HTML助手类-HTML.class.php
类HTML
公共静态输入(类型、id、类、数据、文本)
公共静态标题(模式、文本)
#伪输入标记helper-HTML.class.php::input
函数输入(类型、id、值){
#方法为给定输入创建html字符串。
返回[]。加入(“”);
}
我想这和上面的答案是一样的。但我更喜欢在编写代码时,它应该是模块化的、干净的、漂亮的。始终使用良好的实践。这就是为什么我在这里分享我的想法。如果你创建自己或他人的助手类帮助你更快地发展,我也建议对它的贡献帮助你学习。我知道这不是你想要的答案,但无论如何,你可以随时返回。有效解决方案
<?php
include("database.php");
$db=$conn;
// fetch query
function fetch_data(){
global $db;
$query="SELECT * FROM users WHERE username='vii'"; // change this
$exec=mysqli_query($db, $query);
if(mysqli_num_rows($exec)>0){
$row= mysqli_fetch_all($exec, MYSQLI_ASSOC);
return $row;
}else{
return $row=[];
}
}
$fetchData= fetch_data();
show_data($fetchData);
foreach($fetchData as $data){
$firstname=$data['udid'];} // change this 'udid' to your table field
?>
<!DOCTYPE HTML>
<html>
<head>
<title>Retrieve Contact</title>
</head>
<body>
<input type=text value= <?php echo $firstname; ?>
</body>
</html>
检索联系人
#Pseudo HTML helper class - HTML.class.php
class HTML
public static input(type, id, class, data, text)
public static heading(mode, text)
#Pseudo input tag helper - HTML.class.php::input
function input(type, id, value) {
# method create the html string for the given input.
return ["<input type=",type," id=",id," value=",value,"/>"].join('');
}
<?php
$con = new mysqli(host,user,pass,dbname);
$query = "SELECT * FROM cust WHERE id = '$id'";
$result = $con-> query($query);
while ($row = $result->fetch_assoc()){
$value = $row['id'];
echo HTML::input('text', id, $id);
}
?>
<?php
include("database.php");
$db=$conn;
// fetch query
function fetch_data(){
global $db;
$query="SELECT * FROM users WHERE username='vii'"; // change this
$exec=mysqli_query($db, $query);
if(mysqli_num_rows($exec)>0){
$row= mysqli_fetch_all($exec, MYSQLI_ASSOC);
return $row;
}else{
return $row=[];
}
}
$fetchData= fetch_data();
show_data($fetchData);
foreach($fetchData as $data){
$firstname=$data['udid'];} // change this 'udid' to your table field
?>
<!DOCTYPE HTML>
<html>
<head>
<title>Retrieve Contact</title>
</head>
<body>
<input type=text value= <?php echo $firstname; ?>
</body>
</html>