数据库表中的网站设置

问题描述：

I am building a CMS,

The CMS information is stored in a mysql database inside a table called cms_settings

cms_settings table

I want to get the setting_value and store it to a variable, for example:

$cms_url = ?;
$cms_theme = ?;

I've tried to use the mysqli_fetch_array() but I couldn't make it right.

<?php

require('db.php');

$s = mysqli_query($conn, "SELECT * FROM cms_settings");
$i = mysqli_fetch_array($s);
foreach($i as $key=>$value)
{
  $cms_url = $i[0][2];
}
?>

but that didn't work!

your help is highly appreciated,

我正在构建CMS， p>

CMS信息存储在名为cms_settings code> p>

表 code>中的 mysql code> database code> “https://i.stack.imgur.com/euatp.png"alt =”cms_settings table“> p>

我想获取 setting_value code>并存储它是一个变量，例如： p>

  $ cms_url =？; 
 $ cms_theme =？; 
  code>  pre> 
 
 我尝试使用 mysqli_fetch_array（） code>，但我做不到。 p> 
 
 
 ＆lt;？php 
 
require（  'db.php'）; 
 
 $ s = mysqli_query（$ conn，“SELECT * FROM cms_settings”）; 
 $ i = mysqli_fetch_array（$ s）; 
foreach（$ i as $ key =＆gt; $ 值）
 {
 $ cms_url = $ i [0] [2]; 
} 
？＆gt; 
  code>  pre> 
 
 但这不起作用 ！ p> 
 
 
非常感谢您的帮助， p> 
  div>

答

You need to loop through your results. Maybe try something like this:

<?php

require('db.php');

$settings = array();
$s = mysqli_query($conn, "SELECT * FROM cms_settings");

while($row = mysqli_fetch_assoc($s)) {
      $settings[$row['setting_name']] = $row['setting_value'];
}

edit: added saving your settings into an array.

edit2: If you want to get specific settings, maybe put it in a function.

function getSetting($name) {
    require('db.php');

    $name = mysqli_real_escape_string($conn, $name);

    $s = mysqli_query($conn, "SELECT * FROM cms_settings WHERE `setting_name` = '{$name}'");

    while($row = mysqli_fetch_assoc($s)) {
        return $row['setting_value'];
    }

    return false;
}

答

Its not working because you array is $I and your foreach loop is looping through the $key=>value, but in your foreach you are still referencing $i.

Instead, do this:

while($row = mysqli_fetch_array($s))  {
     $cms_url = $row['cms_url'];
}

EDIT:

A foreach loop in this case would not be ideal as you would battle to give each result a unique variable name as a foreach loop will loop through each array item and apply the required tasks to the current value

答

You can create an array named 'cms_settings' and fill the array with the keys from your result

 <?php
    require('db.php');

    $s = mysqli_query($conn, "SELECT * FROM cms_settings");
    $i = mysqli_fetch_assoc($s);
    $cms_settings= array();

    foreach($i as $row)
    {
      $cms_settings[$row['setting_name']] = $row['setting_value'];
    }

    /* show a setting */
    echo 'CMS url is: '.$cms_settings['cms_url'];

    /* show all settings */
    print_r($cms_settings);

 ?>

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