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php - 如何使变量可从 AJAX/php 查询脚本中单击

转载 作者:行者123 更新时间:2023-11-29 11:22:28 25 4
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在听取了社区的建议后,我开始编写 AJAX/php 脚本,以根据 MySQL 数据库中的变量检索数据。现在它工作得相当好,我希望能够单击显示的任何变量,并将您带到另一个页面,同时记住您的选择。这可能吗?

这是一个link to my website.

这是我的 PokemonFight.php 代码:

<!DOCTYPE html>
<html>
<head>
<script>
function showUser(str) {
if (str == "") {
document.getElementById("txtHint").innerHTML = "";
return;
} else {
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
}
};
xmlhttp.open("GET","getpokemon.php?q="+str,true);
xmlhttp.send();
}
}
</script>
<style>
table {
font-family: arial, sans-serif;
border-collapse: collapse;
width: 100%
}
td, th {
border: 1px solid #dddddd;
text-align: left;
padding: 8px
}
</style>
<title>Pokemon Fight!</title>
</head>
<body>
<h1><center>Pokemon Fight!</center></h1>
<h2>Select your type!</h2>

<form>
<select name="test" onchange="showUser(this.value)">
<option value="">Select a Pokemon Type!:</option>
<option value="1">Normal</option>
<option value="2">Fire</option>
<option value="3">Fighting</option>
<option value="4">Water</option>
<option value="5">Flying</option>
<option value="6">Grass</option>
<option value="7">Poison</option>
<option value="8">Electric</option>
<option value="9">Ground</option>
<option value="10">Psychic</option>
<option value="11">Rock</option>
<option value="12">Ice</option>
<option value="13">Bug</option>
<option value="14">Dragon</option>
<option value="15">Ghost</option>
<option value="16">Dark</option>
<option value="17">Steel</option>
<option value="18">Fairy</option>
</select>
</form>
<br>
<div id="txtHint"><b>Pokemon info will be listed here...</b></div>

</body>
</html>

这是我的 getpokemon.php 代码:

</head>
<body>

<?php
$q = intval($_GET['q']);

$con = mysqli_connect('localhost','root','password','pokemon');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}

mysqli_select_db($con,"pokemon");
$sql="SELECT * FROM test WHERE id = '".$q."'";
$result = mysqli_query($con,$sql);

echo "<table>
<tr>
<th>Pokemon Name</th>
<th>PokedexID</th>
<th>Move 1</th>
<th>Move 2</th>
<th>Move 3</th>
<th>Move 4</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['PokedexID'] . "</td>";
echo "<td>" . $row['Pokemon_Move_1'] . "</td>";
echo "<td>" . $row['Pokemon_Move_2'] . "</td>";
echo "<td>" . $row['Pokemon_Move_3'] . "</td>";
echo "<td>" . $row['Pokemon_Move_4'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
</body>
</html>

最佳答案

下面的代码是我不理解的,但是用这段新代码替换了我之前的代码,在 Pokemon 名称上添加了一个超链接,该链接到我正在使用 PokedexID 的 ID 值创建的另一个页面。希望这对像我一样陷入困境的人有所帮助。

代码之前:

echo "<td>" . $row['PokedexID'] . "</td>"; 

代码后:

echo "<td><a href=' viewpokemon.php?PokedexID=" . $row["PokedexID"] . "'>" . $row["Pokemon_Name"] . "</a></td>"

关于php - 如何使变量可从 AJAX/php 查询脚本中单击,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38690120/

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