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
I'm new here and I have a question, I looked for the solution everywhere and I still cant manage to solve this. 
我是新来的，我有一个问题，我到处都在寻找解决方案，但我仍然无法解决这个问题。
I want to show the results of the SELECT statement (in php) in a table (html) using JS. 我想在使用JS的表（html）中显示SELECT语句（在php中）的结果。 Here is the code of these 3 files: 这是这3个文件的代码：
main.HTML file main.HTML文件
<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 select.PHP文件
<?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 select.js文件
$(document).ready(function(){
$("#button").click(function () {
function show_all() {
$.ajax({
type: "POST",
url: "select.php",
data:{action:"showroom"},
success: function (data) {
$("#id").hide();
$("#content").html(data);
}
});
}
show_all();
});
});
The problem is when I click the button to show the content nothing happens. 问题是当我单击按钮以显示内容时，什么也没发生。
The Select statemnt is correct, in Mysql font I can see the results of the SELECT statement. Select状态是正确的，在Mysql字体中，我可以看到SELECT语句的结果。
This is what I'm getting if I directly run select.php: codnmbredrccn"; while ($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. 该按钮应该在单击时隐藏，但是什么也没有发生。 Seems like it never executes the js file. 似乎它从不执行js文件。
4 个解决方案
解决方案1
0 2015-10-05 09:10:26
In this line 在这条线
echo "<table border='2px'><tr><td>cod</td><td>nmbre</td><td>drccn</td</tr>";
The final </td> is missing a > 最后的</td>缺少>
解决方案2
0 2015-10-05 09:11:39
Everything is perfect in your script if you are sure about your mysql details. 如果您确定自己的mysql详细信息，那么一切在您的脚本中都是完美的。 Please check your js files getting included in script. 请检查您的js文件是否包含在脚本中。
<script src="http://code.jquery.com/jquery-2.1.4.min.js"></script>
解决方案3
0 2015-10-05 10:02:16
I removed jquery.js and inserted another script at head. 我删除了jquery.js并在头插入了另一个脚本。 It worked for me. 它为我工作。 I connected it to localhost. 我将其连接到本地主机。 check it. 核实。
main.html 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 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 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 输出量
解决方案4
-1 2015-10-05 09:41:20
try this 尝试这个
while ($row = mysqli_fetch_assoc($show)) {
echo "<tr><td>" .$row['cod']."</td><td>".$row['nmbre']."</td><td>".$row['drccn']."</td></tr>";
}
or 要么
while ($row = mysqli_fetch_assoc($show)) {
echo "<tr><td>" .$row['0']."</td><td>".$row['1']."</td><td>".$row['2']."</td></tr>";
}
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