PHP array_search无法正常工作 [英] PHP array_search not working

问题描述

有人可以让我知道为什么array_search对我不起作用吗?我想要的就是搜索值并获取相应的键值例如，如果我搜索wiliam，我应该得到4.

can anybody let me know why array_search doesnt works for me? All i want is to search value and and get corresponding key value for eg if i search wiliam i should get 4.. Its simple but aint working for me

<?php
$fqlResult[0]['uid']='1';
$fqlResult[0]['name']='Jay';
$fqlResult[1]['uid']='2';
$fqlResult[1]['name']='UserName2';
$fqlResult[2]['uid']='3';
$fqlResult[2]['name']='Frances';
$fqlResult[3]['uid']='4';
$fqlResult[3]['name']='William';



        for($i=0;$i<count($fqlResult);$i++)
        {

            $userdbname="'".$fqlResult[$i]['name']."'";
            $userdb[$userdbname]="'".$fqlResult[$i]['uid']."'"; 

        }


echo "<pre>";
print_r($userdb);
echo "</pre>";
echo array_search('4', $userdb);
?>

推荐答案

它不起作用，因为 array_seach 搜索值并且"William"是关键.要使事情复杂化，在 for 循环期间，您的值和键会用单引号引起来.

It doesn't work because array_seach searches values and "William" is a key. To complicate things, your values and keys are wrapped in single quotes during the for loop.

您想要执行以下操作:

if ( ! empty($userdb["'William'"]) )
{
  // Echoes "'4'"
  echo $userdb["'William'"];
}

// To find user ID "'4'"
// Outputs "'William'"
echo array_search("'4'", $userdb);

如果您不希望东西用单引号引起来，则需要更改 for 循环，如下所示:

If you don't want things wrapped in single quotes, you'll need to change your for loop as follows:

for($i=0;$i<count($fqlResult);$i++)
{
  $userdbname=$fqlResult[$i]['name'];
  $userdb[$userdbname]=$fqlResult[$i]['uid']; 
}

if ( ! empty($userdb["William"]) )
{
  // Outputs "4" (without the single quotes)
  echo $userdb["William"];
}

// To find user ID "4" (without the single quotes)
// Outputs "William"
echo array_search('4', $userdb);

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