使用php在html表中显示mysql行内容 [英] Showing mysql rows content in html table using php
问题描述
我是新来的,我有一个问题,我到处寻找解决方案,但仍然无法解决这个问题。
我想显示结果在使用JS的表(html)中的SELECT语句(在PHP中)。以下是这3个文件的代码:
main.HTML文件
< HTML>
< head>
< meta charset =UTF-8/>
< script type =text / javascriptsrc =jquery.js>< / script>
< script src =select.jstype =text / javascript>< / script>
< / head>
< body>
< button id =button>显示器< /按钮>
< br>
< input type =textid =id/>
< div id =content>< / div>
< / body>
< / html>
select.PHP文件
<?php
$ link = mysqli_connect(xxxxxxx,user_tienda,%%%%%%%%%,pool_tiendas) ;
if(mysqli_connect_errno())
echoFallo en la conexion con mysql.mysqli_connect_error();
$ action = $ _ POST [action];
if($ action ==showroom){
$ query =SELECT cod,nmbre,drccn from tienda;
$ show = mysqli_query($ link,$ query)或die(error);
echo< table border ='2px'>< tr>< td> cod< / td>< td> nmbre< / td>< td> drccn< / td< ;
while($ row = mysqli_fetch_array($ show)){
echo< tr>< td> 。$行[ '鳕鱼'] < / TD>< TD> 中。$行[ 'nmbre']。 < / TD>< TD> 中$行[ 'drccn']。 < / TD>< / TR>中;
}
echo< / table>;
}
?>
select.JS文件
$(document).ready(function(){
$(#button)。click(function(){
函数show_all(){
$ .ajax({
类型:POST,
url:select.php,
data:{action:showroom} ,
success:function(data){
$(#id)。hide();
$(#content)。html(data);
}
});
}
show_all();
});
});
问题是当我点击按钮来显示内容时什么也没有发生。
选择statemnt是正确的,在Mysql字体中我可以看到SELECT语句的结果。
这是我得到的,如果我直接运行select.php:
codnmbredrccn; while($ row = mysqli_fetch_array($ show)){echo。$ row ['cod' ]。。$ row ['nmbre']。。$ row ['drccn']。;} echo;}?>
现在在mozilla控制台中,我可以看到2个错误:没有se encuentra elemento select.php:18和没有se encuentra elemento main.html:18(没有se encuentra elemento - >无法找到元素
该按钮应该在点击时隐藏,但没有任何反应,似乎它永远不会执行js文件。
我删除了jquery.js,并在头部插入了另一个脚本,它对我有用
我将它连接到本地主机检查它
> main.html
< html>
<头>
< met charset =UTF-8/>
< script src =https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js><<< ; /脚本>
< script src =select.jstype =text / javascript>< / script>
< / head>
< body>
< button id =button>显示器< /按钮>
< br>
< input type =textid =id/>
< div id =content>< / div>
< / body>
< / html>
select.js
$(document).ready(function(){
$(#button)。click(function(){
函数show_all(){
$ .ajax({
类型:POST,
url:select.php,
data:{action:showroom},
success:function(data){
$(#id)。hide();
$(#content)。html(data);
}
});
}
show_all();
});
});
select.php
<?php
$ link = mysqli_connect(localhost,root,,manishYii);
if(mysqli_connect_errno())
echoFallo en la conexion con mysql.mysqli_connect_error();
$ action = $ _ POST [action];
if($ action ==showroom){
$ query =从成员中选择名字,姓氏,EmailID;
$ show = mysqli_query($ link,$ query)或die(error);
echo< table border ='2px'>< tr>< td> cod< / td>< td> nmbre< / td>< td> drccn< / td< ;
while($ row = mysqli_fetch_array($ show)){
echo< tr>< td> 。$行[ '姓'] < / TD>< TD> 中。$行[ '名字'] < / TD>< TD> 中。$行[ 'EMAILID']。 < / TD>< / TR>中;
}
echo< / table>;
}
?>
输出
I'm new here and I have a question, I looked for the solution everywhere and I still cant manage to solve this.
I want to show the results of the SELECT statement (in php) in a table (html) using JS. Here is the code of these 3 files:
main.HTML file
<html>
<head>
<meta charset="UTF-8"/>
<script type = "text/javascript" src="jquery.js" ></script>
<script src="select.js" type="text/javascript"></script>
</head>
<body>
<button id="button"> Mostrar </button>
<br>
<input type="text" id="id" />
<div id="content"></div>
</body>
</html>
select.PHP file
<?php
$link=mysqli_connect("xxxxxxx", "user_tienda","%%%%%%%%","pool_tiendas");
if (mysqli_connect_errno() )
echo "Fallo en la conexion con mysql" .mysqli_connect_error();
$action=$_POST["action"];
if ($action=="showroom") {
$query = "SELECT cod, nmbre, drccn from tienda";
$show = mysqli_query($link, $query) or die ("error");
echo "<table border='2px'><tr><td>cod</td><td>nmbre</td><td>drccn</td</tr>";
while ($row = mysqli_fetch_array($show)) {
echo "<tr><td>" .$row['cod']."</td><td>".$row['nmbre']."</td><td>".$row['drccn']."</td></tr>";
}
echo "</table>";
}
?>
select.JS file
$(document).ready(function(){
$("#button").click(function () {
function show_all() {
$.ajax({
type: "POST",
url: "select.php",
data:{action:"showroom"},
success: function (data) {
$("#id").hide();
$("#content").html(data);
}
});
}
show_all();
});
});
The problem is when I click the button to show the content nothing happens.
The Select statemnt is correct, in Mysql font I can see the results of the SELECT statement.
This is what I'm getting if I directly run select.php: codnmbredrccn"; while ($row = mysqli_fetch_array($show)) { echo "" .$row['cod']."".$row['nmbre']."".$row['drccn'].""; } echo ""; } ?>
Now in mozilla console I can see 2 errors: no se encuentra elemento select.php:18 and no se encuentra elemento main.html:18 (no se encuentra elemento -> can't find element
The button is supposed to hide when clicked, but nothing happens. Seems like it never executes the js file.
I removed jquery.js and inserted another script at head. It worked for me. I connected it to localhost. check it.
main.html
<html>
<head>
<meta charset="UTF-8"/>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script src="select.js" type="text/javascript"></script>
</head>
<body>
<button id="button"> Mostrar </button>
<br>
<input type="text" id="id" />
<div id="content"></div>
</body>
</html>
select.js
$(document).ready(function(){
$("#button").click(function () {
function show_all() {
$.ajax({
type: "POST",
url: "select.php",
data:{action:"showroom"},
success: function (data) {
$("#id").hide();
$("#content").html(data);
}
});
}
show_all();
});
});
select.php
<?php
$link=mysqli_connect("localhost", "root","","manishYii");
if (mysqli_connect_errno() )
echo "Fallo en la conexion con mysql" .mysqli_connect_error();
$action=$_POST["action"];
if ($action=="showroom") {
$query = "SELECT FirstName, LastName, EmailID from members";
$show = mysqli_query($link, $query) or die ("error");
echo "<table border='2px'><tr><td>cod</td><td>nmbre</td><td>drccn</td</tr>";
while ($row = mysqli_fetch_array($show)) {
echo "<tr><td>" .$row['FirstName']."</td><td>".$row['LastName']."</td><td>".$row['EmailID']."</td></tr>";
}
echo "</table>";
}
?>
Output
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