所以我正在尝试从头开始创建php方法。我的课程还不完全是课程,我仍在努力。无论如何,我的问题是我似乎无法从数据库中获得我期望的值。这是我的代码片段:
文件1.php
<?php function dbConnect() {
$connection = mysqli_connect("localhost", "music_root", "", "music");
if($connection->connect_error) {
return null;
}
return $connection;}
function getCategory() {
$cn = dbConnect();
if($cn == null) {
echo "Failed to connect to database.";
} else {
$fetchQuery = mysqli_query($cn, "SELECT * FROM tbl1 ORDER BY 'Name'") or die(mysqli_error($cn));
if(mysqli_num_rows($fetchQuery) > 0) {
while($item = mysqli_fetch_array($fetchQuery)) {
return $item["Name"];
}
}
}} ?>
这是我如何在 file2 中调用上述方法的代码片段.php
<?php ini_set("display_errors", 1);
include_once("file1.php");
$con = dbConnect();
$updateStat = false; ?>
<div>
<label>Genre</label>
<select id="genre" name="genre" value="Please select genre">
<option value="<?php $con->getCategory() ?>"></option>
</select>
</div>
我尝试在方法开始时打印一条消息,以查看它是否被调用,但消息也没有打印,所以我想知道,我在这里可能错过了什么?
我认为您的代码中有服务器错误...我想,您不使用OOP(类),所以我修改了一个应该可以工作的示例..如果没有,请发布错误消息
文件1.php
function getCategory($cn) {
$out = array();
if($cn == null) {
echo "Failed to connect to database.";
} else {
$fetchQuery = mysqli_query($cn, "SELECT * FROM tbl1 ORDER BY 'Name'") or die(mysqli_error($cn));
if(mysqli_num_rows($fetchQuery) > 0) {
while($item = mysqli_fetch_array($fetchQuery)) {
$out[] = $item["Name"];
}
}
return $out;
}
}
fil2.php
<?php
ini_set("display_errors", 1);
require_once("file1.php");
$con = dbConnect();
$updateStat = false;
$res = getCategory($con);
?>
<div>
<label>Genre</label>
<select id="genre" name="genre" value="Please select genre">
<?php
foreach($res as $cat):
?>
<option value="<?php echo $cat ?>"><?php echo $cat ?></option>
<?php endforeach;?>
</select>