我有这两个表“fcategory”和“fthreads”。 fcategory 字段:category_id、category_name。 fthreads 字段:thread_id、thread_title、category_name、category_id、user_id。
当我在 php 中创建一个新线程时,我希望 category_id 与 category_name 一起获取并将其插入到 fthreads 表中。
这是我的 php 文件:threads.php
<form action="threadsp.php" name="myform" method="post">
<label for="field4"><span>Category</span>
<?php
$query = "select * from fcategory";
$result = mysqli_query($conn, $query);
$resultsearch = mysqli_fetch_array($result);
if(!$result){
die('could not get data:'.mysqli_error($conn));
}
echo '<select name="category" class="select-field">';
while ($row = mysqli_fetch_assoc($result)) {
echo '<option value="'.$row['category_name'].'">'.$row['category_name'].'</option>';
}
echo "</select>";
echo "</label>";
?>
<label for="field1"><span>Thread Title <span class="required">*</span></span>
<input type="text" class="input-field" name="title" value="" />
</label>
<label><span> </span>
<input type="submit" value="Create" />
</label>
</form>
第二个文件:threadsp.php
<?php
$catg = $_POST['category'];
$title = $_POST['title'];
$userid = $_SESSION['uid'];
$sql = "select category_id from fcategory where category_name = '$catg'";
$result2 = mysqli_query($conn, $sql);
$catgidresult = mysqli_fetch_array($result2);
$query = "insert into fthreads (category_name,thread_title,user_id,category_id) values('".$catg."','".$title."','".$userid."','".$catgidresult."')";
$result = mysqli_query($conn, $query);
if(!$result){
echo "failed".mysqli_error($conn);
}else{
header("Location: question.php");
die();
}
?>
我能够获取 category_name 但 category_id 的值显示为 0。 任何帮助将不胜感激。谢谢
最佳答案
为什么要将 category_name
存储在 fthreads 表
中,如果该表中有 category_id ?只需从您的 fcategory 表
中删除 category_name 即可。您需要规范化您的表格。参见 this :所以,只需从您的表单发送 category_id
,如下所示:
<form action="threadsp.php" name="myform" method="post">
<label for="field4"><span>Category</span>
<?php
$query = "select * from fcategory";
$result = mysqli_query($conn, $query);
$resultsearch = mysqli_fetch_array($result);
if(!$result){
die('could not get data:'.mysqli_error($conn));
}
echo '<select name="cat_id" class="select-field">';
while ($row = mysqli_fetch_assoc($result)) {
echo '<option value="'.$row['category_id'].'">'.$row['category_name'].'</option>';
}
echo "</select>";
echo "</label>";
?>
<label for="field1"><span>Thread Title <span class="required">*</span></span>
<input type="text" class="input-field" name="title" value="" />
</label>
<label><span> </span>
<input type="submit" value="Create" />
</label>
</form>
在你的 threads.php 文件中你可以这样做:
<?php
$cat_id = $_POST['cat_id'];
$title = $_POST['title'];
$userid = $_SESSION['uid'];
$query = "insert into fthreads (thread_title,user_id,cat_id) values('".$title."','".$userid."','".$cat_id."')";
$result = mysqli_query($conn, $query);
if(!$result){
echo "failed".mysqli_error($conn);
}else{
header("Location: question.php");
die();
}
?>
此外,您需要清理您的输入,否则您很容易受到 Sql 注入(inject)攻击。参见 here是清理输入的最佳方法。 快乐编码 :) .
关于Php:在执行插入查询时检索一个表的主键并存储到另一个表中。,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35217698/