我在这里有mysql记录显示在带有删除按钮的html表中。我需要做的是禁用删除按钮,如果两个数据库表中都存在记录。
如果两个表中都已存在记录,如何禁用每行的删除按钮? 任何帮助将不胜感激。
$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>";
}
抛出一个简单的 if() 语句与两个查询连接在 pdo 中,您可以使用 ->rowCount()不确定在 mysqli 中
所以你需要这个逻辑
查询 1 = 表 1 中的计数行
query2 = 表 2 中的计数行
就好像你说如果它存在于两个表中,它应该隐藏它,所以你将使用一个if-or statement
if(query1 == 0 || query2 == 0){
//show your button
}
这里的立场很简单:
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
如果您希望我提供PDO示例,请回复
编辑:
PDO类,使其更容易连接
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:更新了 &&和 2x query1 检查的小错误
$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;)如果我是对的,您可以从 sql 查询中的多个表中进行检查,因此如果您进行以下 1 个查询的多重检查,您可以使用 1 个查询来完成!
编辑:这个问题的逻辑步骤:
1) Connect
database
2)为table1
做一个query
3) 从table1
Count
entries
(records
4)为table2
做一个query
5) 从table2
Count
entries
(records
6)Check
如果其中one
等于0
7)如果其中one
等于0
entries
(records
(rows
)),则显示按钮