我这里的mysql记录显示在带有删除按钮的html表中。我需要做的是,如果两个数据库表中都有记录,则禁用删除按钮。
如果两个表中都已存在记录,如何禁用每行删除按钮?任何帮助都会表示赞赏。
$search = $mysqli1->real_escape_string($_POST['bid']);
$search = preg_replace("/[^A-Za-z0-9 ]/", '', $search);
$search = $_POST['bid'];
$res = $mysqli1->query("select * from code WHERE item LIKE '%$search%' OR item_code LIKE '%$search%' OR cat_code LIKE '%$search%' order by item_code ASC");
while($r = $res->fetch_assoc()){
echo "<tr>
<td><a href='#' id='".$r['id']."' class='del'><img src='../images/del.png'></a></td>
</tr>";
}
答案 0 :(得分:0)
抛出一个简单的if()语句 与两个查询相关联 在pdo中你使用 - &gt; rowCount() 在mysqli中不确定
所以你需要这个逻辑
query1 =计算table1中的行
query2 =计算table2中的行
好像你说两个表中是否存在它应该隐藏它所以你将使用if-or statement
if(query1 == 0 || query2 == 0){
//show your button
}
这里的内容很简单:
if(query1 equals 0 rows OR query2 equals 0 rows){
//show your button
}
//While you don't put up the else with something else it won't show anything
//so if there are the value of 1+ rows in both query1 and query2 this won't show anything
如果你想让我提供一个pdo示例,只需回复
编辑:
PDO Class,使连接更容易
class Database extends PDO
{
private $db;
public function Database($host, $user, $pass, $db) {
try {
$this->db = new PDO("mysql:dbname=".$db.";host=".$host.";", $user, $pass);
} catch(PDOEXCEPTION $e) {
die('An error has occurred! [Code: '.$e->getCode().']! <br/>More info: ['.$e->getMessage().']!');
}
}
public function runQuery($query) {
try{
return $this->db->query($query);
} catch(PDOEXCEPTION $e) {
die('An error has occurred! [Code: '.$e->getCode().']!<br/>More info: ['.$e->getMessage().']!');
}
}
}
现在行计数:已更新&amp;&amp;&amp;和2x query1检查
$consite = new Database('DBHost','DBUsername','DBPassword','DBName');
$query1 = $consite->runQuery("SELECT * FROM TABLE1");
$query2 = $consite->runQuery("SELECT * FROM TABLE2");
if($query1->rowCount() == 0 || $query2->rowCount() == 0) {
//do your while statement to loop through it
//if you done your while statement it only shows the delete button
//for items that are NOT in both tables
}
抱歉,我懒得添加while语句;)
如果我是正确的,你可以在sql查询中检查多个表,这样你就可以在1次查询后进行多次检查,从而可以进行1次查询!
编辑: 这个问题的逻辑步骤:
1)Connect
到database
2)为query
table1
3)Count
来自entries
的{{1}}(records
}
4)为table1
query
5)table2
来自Count
的{{1}}(entries
}
6)records
如果table2
等于Check
7)如果one
等于0
one
(0
(entries
)),则显示按钮