PHP - 让我查询的数组键ID [英] PHP - Make my query's array's key the ID

问题描述

所以我有这个阵列中拔出图片和数组的键只是0,1,2,3,4,5 ....和诸如此类的东西...

我怎样才能使该表中的关键的ID列中的值，并保持链接作为值。

关联数组，不是吗？

这是我的PHP：

  $ myImageID = $我['imageid'];$ findImages =SELECT FROM WHERE图像= MODEL_ID'{$我['身份证']}链接;
$ imageResult =的mysql_query（$ findImages）或死亡（mysql_error（））;$ myImages =阵列（）;
而（$行= mysql_fetch_array（$ imageResult））{
    $ myImages [] = $行[0];
}
 

这是我有：

  {
 [0]  -  GT; http://website.com/link.jpg
 [1]  - ＆GT; http://website.com/li123nk.jpg
 [2]  - ＆GT; http://website.com/link3123.jpg
}
 

这就是我想要的：

  {
 [47]  -  GT; http://website.com/link.jpg
 [122]  -  GT; http://website.com/li123nk.jpg
 [4339]  -  GT; http://website.com/link3123.jpg
}
 

解决方案

只需选择ID，并使其成为关键阵列。就这么简单。

  $ findImages =SELECT ID，从图像中的WHERE MODEL_ID ='{$我['身份证']}链接;
$ imageResult =的mysql_query（$ findImages）或死亡（mysql_error（））;$ myImages =阵列（）;
而（$行= mysql_fetch_array（$ imageResult））{
    $ myImages [$行[0] = $行[1];
}
 

仅供参考，你不应该在新的$使用的mysql _ * 功能C $ç。他们不再维护并正式去precated 。请参阅红色框？了解的 prepared语句的，而是和使用的 PDO 或的MySQLi - 的这篇文章将帮助您决定哪。如果您选择PDO，是一个很好的教程。

您还敞开 SQL注入

So I have this array of images pulling and the array's keys are just 0,1,2,3,4,5.... and whatnot...

How can I make the value in the 'id' column of that table the key, and keep 'link' as the value.

Associative Array, no?

Here is my PHP:

$myImageID = $me['imageid'];

$findImages = "SELECT link FROM images WHERE model_id ='{$me['id']}'";
$imageResult = mysql_query($findImages) or die (mysql_error());

$myImages = array();
while($row = mysql_fetch_array($imageResult)) {
    $myImages[] = $row[0];
}

Here is what I have:

{
 [0] -> "http://website.com/link.jpg"
 [1] -> "http://website.com/li123nk.jpg"
 [2] -> "http://website.com/link3123.jpg"
}

Here is what I want:

{
 [47] -> "http://website.com/link.jpg"
 [122] -> "http://website.com/li123nk.jpg"
 [4339] -> "http://website.com/link3123.jpg"
}

解决方案

Just select the id and make it the key to the array. It's that simple.

$findImages = "SELECT id, link FROM images WHERE model_id ='{$me['id']}'";
$imageResult = mysql_query($findImages) or die (mysql_error());

$myImages = array();
while($row = mysql_fetch_array($imageResult)) {
    $myImages[$row[0]] = $row[1];
}

FYI, you shouldn't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial.

You also wide open to SQL injections

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