我有这个脚本只允许用户输入单个标签,但我想让用户输入多个用逗号分隔的标签,例如shoe, shirt, hat, glasses
,并将每个标签存储在数据库中。
有人可以给我一些我需要在脚本中更改的示例,以便执行此操作。
以下是我的MySQL表格。
CREATE TABLE questions_tags (
id INT UNSIGNED NOT NULL AUTO_INCREMENT,
tag_id INT UNSIGNED NOT NULL,
users_questions_id INT UNSIGNED NOT NULL,
PRIMARY KEY (id)
);
CREATE TABLE tags (
id INT UNSIGNED NOT NULL AUTO_INCREMENT,
tag VARCHAR(255) NOT NULL,
PRIMARY KEY (id)
);
以下是以下脚本。
<?php
require_once ('./mysqli_connect.php');
if (isset($_POST['submitted'])) {
$mysqli = new mysqli("localhost", "root", "", "sitename");
$dbc = mysqli_query($mysqli,"SELECT questions_tags.*, tags.* FROM questions_tags, tags");
if (!$dbc) {
print mysqli_error($mysqli);
}
$page = '3';
$tag = mysqli_real_escape_string($mysqli, $_POST['tag']);
$mysqli = new mysqli("localhost", "root", "", "sitename");
$dbc = mysqli_query($mysqli,"SELECT questions_tags.*, tags.* FROM questions_tags INNER JOIN tags ON tags.id = questions_tags.tag_id WHERE questions_tags.users_questions_id='$page'");
if(mysqli_num_rows($dbc) >= 0){
$mysqli = new mysqli("localhost", "root", "", "sitename");
$clean_url = mysqli_real_escape_string($mysqli, $page);
$query1 = "INSERT INTO tags (tag) VALUES ('$tag')";
if (!mysqli_query($mysqli, $query1)) {
print mysqli_error($mysqli);
return;
}
$mysqli = new mysqli("localhost", "root", "", "sitename");
$dbc = mysqli_query($mysqli,"SELECT id FROM tags WHERE tag='$tag'");
if (!$dbc) {
print mysqli_error($mysqli);
} else {
while($row = mysqli_fetch_array($dbc)){
$id = $row["id"];
}
}
$query2 = "INSERT INTO questions_tags (tag_id, users_questions_id) VALUES ('$id', '$page')";
if (!mysqli_query($mysqli, $query2)) {
print mysqli_error($mysqli);
return;
}
echo "$tag has been entered";
if (!$dbc) {
print mysqli_error($mysqli);
}
}
mysqli_close($mysqli);
}
?>
答案 0 :(得分:2)
您可以使用explode()
获取以逗号分隔的标签数组
$tag_string = "t1, t2, t3";
$tags = explode(",", $tag_string );
echo $tags[0]; // t1
echo $tags[1]; // t2
然后你可以循环遍历数组以插入数据库
您可能还希望创建查询包含UNIQUE
CREATE TABLE tags (
id INT UNSIGNED NOT NULL AUTO_INCREMENT,
tag VARCHAR(255) NOT NULL,
PRIMARY KEY (id),
UNIQUE(`tag`)
);
这样你就不会有两个同名的标签。请在此处查看有关UNIQUE语法
的进一步说明此处进行编码而不测试xD
//Assuming you have already added the question and the mysql_insert_Id() == 1
//where mysql_insert_Id() is the last id added to the question table
if (isset($_POST['tags'])){
$tags = explode(",", $_POST['tags']);
for ($x = 0; $x < count($tags); $x++){
//Due to unique it will only insert if the tag dosent already exist
mysql_query("INSERT INTO tag VALUES(NULL, {$tags[x]})");
//Add the relational Link
mysql_query("INSERT INTO question_tag VALUES(NULL, (SELECT tags.Id FROM tags WHERE tags.tag = {$tags[x]}), 1)");
}
}
答案 1 :(得分:1)
避免使用子查询,mysql_insert_id()
为实现相同目标提供了良好的结果。
另一件事 - 在$_POST['tags']
之前检查explode()
是否有逗号以确保您将获得一个数组,并检查循环是否为trim($tags[$x]) == ''
(什么如果$ _POST ['tags']可能会发生,例如:$_POST['tags'] === "tag1,"
或"tag1, "
。
并没有与问题本身联系,但尝试每个请求只使用一个数据库连接(只要没有充分的理由以其他方式执行)。