如何在数组中创建 href link

Question

实际上，我必须使用 php 数组将 table 数据填充到我的仪表板。 table 包含 URL。我需要设置用于重定向到动态 link 的 href。（下面是我的 while 循环代码）。我尝试使用下面的单行 echo php code.It 创建 href 工作并且我得到了可点击的 link，但它没有 return 数据在 table.它显示了 table 中的数据，没有任何对齐。

echo '<html><head></head><a href="'.$data['link'].'" target="_blank">'.$data['link'].'</a></html>';

但我想在数组中创建 href 请任何人帮助我们。

public function getTemperatures()
{
    $mysql_hostname = "localhost";
    $mysql_user     = "root";
    $mysql_password = "";
    $mysql_database = "new";
    $bd = mysql_connect($mysql_hostname, $mysql_user, $mysql_password) or die("Oops some thing went wrong");
    mysql_select_db($mysql_database, $bd) or die("Oops some thing went wrong");// we are now connected to database

    $result = mysql_query("SELECT * FROM website"); // selecting data through mysql_query()
    $pie = mysql_query("select count(*) from website where status = 'Error'");

    while($data = mysql_fetch_array($result))
    {
        //echo '<html><head></head><a href="'.$data['link'].'" target="_blank">'.$data['link'].'</a></html>';
        $temperatures[] = array(
            'label1' => $data['link'],
            'label2' => $data['time'],
            'label3' => $data['os'],
            'label4' => $data['browser'],
            'label5' => $data['status'],
            'label6' => $data['location'],
            array('label6' => $data['widget_load_time'])
        );
    }    
    return DataTable::makeFromIndexedArray($temperatures);
}

Answer 1

看来您已经有了解决方案。只需更改将 'label1' 分配给以下内容的方式即可。

public function getTemperatures()
{
    $mysql_hostname = "localhost";
    $mysql_user     = "root";
    $mysql_password = "";
    $mysql_database = "new";
    $bd =  mysql_connect($mysql_hostname, $mysql_user, $mysql_password) or die("Oops some thing went wrong");
    mysql_select_db($mysql_database, $bd) or die("Oops some thing went wrong");// we are now connected to database

    $result = mysql_query("SELECT * FROM website"); // selecting data through mysql_query()
     $pie = mysql_query("select count(*) from website where status = 'Error'");

    while($data = mysql_fetch_array($result))
    {
        $temperatures[] = array(
            'label1' => '<a href="'.$data['link'].'" target="_blank">'.$data['link'].'</a>',
            'label2' => $data['time'],
            'label3' => $data['os'],
            'label4' => $data['browser'],
            'label5' => $data['status'],
            'label6' => $data['location'],
            array('label6' => $data['widget_load_time'])
        );
    }    
    return DataTable::makeFromIndexedArray($temperatures);
}

如评论所述，避免使用 mysql_* 函数，因为它们不再受支持（这应该是足够的理由）

Answer 2

请添加这一行

'label1' => '<a href="'.$data['link'].'" target="_blank">'.$data['link'].'</a>',

如何在数组中创建 href link

How to create href link inside array

php

mysql

matomo