Question

我创建了一个数组，其中一个数组是php用来显示从sqlite3检索到的记录中的字段的字符串。

我的问题是......它没有。

定义数组，＆＃34; 1＆＃34;作为第一个数据库领域，＆＃34; 2＆＃34;是第二个数据库字段：

编辑：我已将问题重新定义为脚本，以便您可以看到整个内容：

//If I have an array (simulating a record retrieved from database):
$record = array(
    name => 'Joe',
    comments => 'Good Bloke',
    );

//then I define an array to reference it:
$fields = array( 
    1 => array(
    'db_index' => 'name',
    'db_type' => 'TEXT',
    'display' => '$record["name"]',
    'form_label' => 'Name',
    ),
    2 => array(
    'db_index' => 'comments',
    'db_type' => 'TEXT',
    'display' => '$record["comments"]',
    'form_label' => 'Comments',
    ),  
);  

//If I use the lines:
print "expected output:\n";
print " Name = " . $record["name"] ."\n";
print " Comments = " . $record["comments"] ."\n";

//I display the results from the $record array correctly.
//However if I try & use the fields array with something like:
Print "Output using Array values\n";
foreach($GLOBALS["fields"] as $field)
    {
        $label = $field['form_label'];
        $it = $field['display'];    
        $line = "\"$label = \" . $it .\"\n\"";
        print $line;
    }

输出：

Expected output:
 Name = Joe
 Comments = Good Bloke
Output using Array values:
 Name = $record["name"] 
 Comments = $record["comments"]

Answer 1

不要从字符串中调用变量。只是连接它：

foreach($GLOBALS["fields"] as $field){
    $label = $field['form_label'];
    $it = $field['display'];

    eval("$it = ".$it);
    $line = $label." = ".$it."\n";
    print $line;
}

嗯，看起来怎么样？

使用数组值从另一个数组中检索

1 个答案: