[英]Insert Multiple sql statements in php from an array from for loop or foreach
[英]Capturing/INSERT specific value from an input element added to a foreach loop that uses a SQL query array
所以,我一直在研究这个问题大约一个星期。 我发现了很多类似的问题,有类似this one或this one 的解决方案,但我对它们的运气并不好。
我的问题是,我试图从与 foreach 循环返回的结果相关联的任何一个输入元素中捕获特定值。 像搜索引擎一样,提交查询,返回结果,每个结果都添加了一个“评级”滑块,用户应该能够对任何结果进行评级并提交该结果的唯一评级。
我的“评级”范围输入对象:
<form action='#' method='POST'>
<input class='rating' type='range' name='rating_s' min='0' max='100' step='1' value='50' /></i>
<input class="rateglowbutton" type=submit value=Rate />
</form>
它位于 foreach 循环中(我知道,有很多用于格式化的回声,就是这样):
// array
$audios = $wpdb->get_results($wpdb->prepare("
SELECT a.aud_id
, a.date
, a.teacher
, a.title
, a.teach_desc
, a.venue
, a.genre
, a.url
, GROUP_CONCAT(t.tag_name SEPARATOR ', ') tag_name
FROM audio_tag_xref atx
RIGHT
JOIN audios a
ON atx.aud_id = a.aud_id
LEFT
JOIN audio_tags t
ON atx.tag_id = t.tag_id
WHERE (a.title LIKE '%$strKey%'
OR a.venue LIKE '%$strKey%'
OR a.teach_desc LIKE '%$strKey%'
OR a.genre LIKE '%$strKey%')
GROUP
BY a.aud_id
ORDER
BY a.aud_id DESC
",$strKey));
<?php echo "<table>"; ?>
<?php foreach($audios as $i => $audio){ ?>
<?php echo "<tr>";
echo "<tr><h3>".$audio->title."</h3></tr>";
echo "<tr>".$audio->url."</tr>";
echo "</br>";
echo "<tr>".$audio->aud_id."</tr>";
echo "</br>";
echo "<tr><b>".$audio->teacher."</b></tr>";
echo "</br>";
echo "<tr><i>".$audio->date."</i></tr>";
echo "  |  ";
echo "<tr><i>".$audio->venue."</i></tr>";
echo "</br>";
echo "<tr>".$audio->genre."</tr>";
echo "</br>";
echo "<tr>".$audio->teach_desc."</tr>";
echo "</tr>";
echo "</br>";
echo "</br>"; ?>
<form action='#' method='POST'>
<input class='rating' type='range' name='rating_s[]' min='0' max='100' step='1' value='50' />  <i class="fa fa-plus" aria-hidden="true"></i></div>
<input class="rateglowbutton" type=submit value=Rate />
</form>
<?php } ?>
<?php echo "</table>"; ?>
然后将提交的评分插入到我的数据库中的评分表中:
<?php
if (isset($_POST['rating_s'])) {
$userid = md5($_SERVER['REMOTE_ADDR']);
$rate = $_POST['rating_s'];
$aud_id = "$audio->aud_id";
$sql = $wpdb->query($wpdb->prepare("INSERT INTO ratings (rate, aud_id, user_id) VALUES ('$rate', '$aud_id', '$userid')",$rate,$aud_id,$userid));
if ($sql) {
echo "</br>";
echo "<div class='rate_msg_cont'><div class='rate_msg'>You gave a rating of $rate for teaching ID: $aud_id.</div></div>";
}
}
?>
但是,我无法同时将费率和 aud_id 正确插入到评分表中。 充其量,我能够获得评级,但无法获得它在页面上“属于”的 aud_id(为其各自返回的 $audio 结果呈现的输入对象)。 我试图遵循:
这是没有在 foreach 中正确索引输入元素的问题吗? 也许是处理数组的问题? 完全是别的什么?
总结一下:我有一个搜索页面,它使用输入文本框供用户输入关键字字符串。 该字符串返回一个数组,foreach 循环使用该数组来输出查询结果,其中包括一个允许用户对音频进行评级的输入对象。 可以很好地提交费率,但它们最终不会与用户实际尝试评价的结果音频相关联。
如何确保在对结果进行评级时,将评级和相应的 aud_id 作为新条目插入到评级表中?
视觉,如果有帮助:
非常感谢任何人的帮助。
首先从远处看这个……我看到你在使用 PDO,但后来我看到你逃避了 PDO 的全部意义。
您不应该将您的值直接回显到您的查询字符串中...您应该绑定它们。
我只想给你一个小的 PDO 例子(与你的没有特别的关系),这样也许你可以用它作为指南来修复你自己的错误。
<table class="table table-bordered table-striped mb-none" id="datatable-editable">
<thead>
<tr>
<th>Accout Type | User's Name | Email Address</th>
<th></th>
</tr>
</thead>
<tbody id="tablebody">
<?
$DbHost = ""; // The host where the MySQL server resides
$DbDatabase = ""; // The database you are going to use
$DbUser = ""; // Username
$DbPassword = ""; // Password
// --- PDO Info
$dsn = 'mysql:host='.$DbHost.';dbname='.$DbDatabase;
$DbOptions = array(PDO::MYSQL_ATTR_INIT_COMMAND => 'SET NAMES utf8',PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION);
$DBH = new PDO($dsn, $DbUser, $DbPassword, $DbOptions);
$qs = "SELECT * FROM `users` WHERE `accountid`=:accountid";
$q = $DBH->prepare($qs);
$q->bindValue(':accountid',(integer)$userrow['accountid'], PDO::PARAM_INT);
$q->execute();
while($row = $q->fetch(PDO::FETCH_ASSOC))
{
if($userrow['access'] >= $row['access'])
{
?>
<tr id="row<?=$row['id'];?>">
<?
$qs2 = "SELECT COUNT(*) AS `count` FROM `accounts` WHERE `admin` = :id";
$q2 = $DBH->prepare($qs2);
$q2->bindValue(':id', (integer)$row['id'], PDO::PARAM_INT);
$q2->execute();
$count = $q2->fetch(PDO::FETCH_ASSOC)['count'];
$isAdmin = 'false';
$fontStyle = " style='color:blue;'";
$userType = "Developer";
if($row['access'] == 40)$userType = "Accountant";
if($row['access'] == 60)$userType = "Administrator";
if($row['access'] == 99)$userType = "Internal Administrator";
if($row['access'] == 100)$userType = "Internal Developer";
$root = "<span class='label label-info'>$userType</span> ";
if($count > 0)
{
$isAdmin = 'true';
$fontStyle = " style='font-weight:bold;color:red;'";
$root = "<span class='label label-danger'>Root</span> ";
}
?>
<td><span style="color:red;"><?=$root;?></span><span<?=$fontStyle;?>><?=ucfirst(strtolower($row['firstname']))." ".ucfirst(strtolower($row['lastname']));?> - <?=$row['email'];?></span></td>
<?
if($count == 0)
{
?>
<td><a class="mb-xs mt-xs mr-xs" href="javascript:edit(<?=$row['id'];?>, '<?=$row['firstname'];?>', '<?=$row['lastname'];?>', '<?=$row['email'];?>', <?=$row['access'];?>, <?=$isAdmin;?>)">Edit</a></td>
<?
}
else
{
?><td>This account cannot be modified here.</td><?
}
?>
</tr>
<?
}
else
{
?>
<tr id="row<?=$row['id'];?>">
<?
$qs2 = "SELECT COUNT(*) AS `count` FROM `accounts` WHERE `admin` = :id";
$q2 = $DBH->prepare($qs2);
$q2->bindValue(':id', (integer)$row['id'], PDO::PARAM_INT);
$q2->execute();
$count = $q2->fetch(PDO::FETCH_ASSOC)['count'];
$isAdmin = 'false';
$fontStyle = " style='color:blue;'";
$root = "<span class='label label-info'>User Account</span> ";
if($count > 0)
{
$isAdmin = 'true';
$fontStyle = " style='font-weight:bold;color:red;'";
$root = "<span class='label label-danger'>Root Account</span> ";
}
?>
<td><span style="color:red;"><?=$root;?></span><span<?=$fontStyle;?>><?=ucfirst(strtolower($row['firstname']))." ".ucfirst(strtolower($row['lastname']));?> - <?=$row['email'];?></span></td>
<td>You do not have access to modify this account.</td>
</tr>
<?
}
}
?>
</tbody>
</table>
好的,感谢这里的每个人的帮助以及一些进一步的输入/试验和错误,我已经找到了一个解决方案。 我仍然会按照建议进行一些清理,但这里的想法和代码使每个输入元素都具有独特的作用:
表单元素修改:
<?php echo " <form action='#' method='POST'>"; ?>
<?php echo " <input type='hidden' name='audio_index' value='" . $i . "' />"; ?>
<?php echo " <input type='hidden' name='audio_id' value='" .$teaching->aud_id. "' />"; ?>
<?php echo " <div class='range_container'>"; ?>
<?php echo " <div class='rate_sym'>"; ?>
<?php echo " <i class='fa fa-minus' aria-hidden='true'></i>  "; ?>
<?php echo " <input class='rating' type='range' name='rating_s' min='0' max='100' step='1' value='50' />  "; ?>
<?php echo " <i class='fa fa-plus' aria-hidden='true'></i>"; ?>
<?php echo " </div>"; ?>
<?php echo " </div>"; ?>
<?php echo " <div class='rate_container'>"; ?>
<?php echo " <div class='ratescr_container'></div>"; ?>
<?php echo " <input class='rateglowbutton' type='submit' value='Rate' />" ?>
<?php echo " </div>"; ?>
<?php echo " </form>"; ?>
echo " </td>";
echo " </tr>"; ?>
POST 代码修改:
<?php
if (isset($_POST['rating_s']) && isset($_POST['audio_id'])) {
$table = "ratings";
$parama = $_POST['audio_id'];
$paramb = $_POST['rating_s'];
$paramc = md5($_SERVER['REMOTE_ADDR']);
$sql = $wpdb->query($wpdb->prepare("INSERT INTO $table (a_id, rating, u_id) VALUES ('$parama', '$paramb', '$paramc')"));
if (query) {
echo "</br>";
echo "<div class='rate_msg_cont'><div class='rate_msg'>Hooray! You rated
audio $parama a $paramb.</div></div>";
}
}
?>
请注意表单中的附加隐藏输入元素,用于为返回的每个音频生成索引,并为要使用的评级输入生成“audio_id”。 同样,我理解可以(将)实施的清理和安全改进,但我想将此作为工作解决方案发布给任何需要提示处理相同或类似问题的人。
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