Question

我想像这样输出

var data = { 
{
"cname": "Albania"
}
{
"cname": "Austria"
}
}

我试过这段代码，但收到错误"Parse error: syntax error, unexpected T_VAR in C:\wamp\www\mvc\map\data.php on line 22"

<?php
header('Content-type: application/json');  // this is the magic that sets responseJSON

 $dbhost = "localhost";
$dbuser = "root";
$dbpass = "";
$dbname = "test";

// Connecting, selecting database
$link = mysql_connect($dbhost, $dbuser, $dbpass)
    or die('Could not connect: ' . mysql_error());
mysql_select_db($dbname) or die('Could not select database');

        $query = "SELECT cname FROM country";       // Performing SQL query
        $result = mysql_query($query) or die('Query failed: ' . mysql_error());
        $all_recs = array();
        while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
            $all_recs[] = $line;
        }


var data = {

echo json_encode($all_recs);

};

// Free resultset
mysql_free_result($result);

// Closing connection
mysql_close($link);
?>

任何人都可以指导我如何解决此问题，谢谢

Answer 1

我不确定你为什么要这样，但是你可以在php中混合使用js代码，你可以试试

$jsonData = json_encode($all_recs);
echo "var data = {$jsonData}";

Answer 2

$data = "cname : Albania cname : Austria";

解析错误：语法错误，意外T_VAR

2 个答案: