2009-09-22 72 views

回答

1
<?php 
$arr = array ('a'=>1,'b'=>2,'c'=>3,'d'=>4,'e'=>5); 
// of course this array can be created by looping through your mysql result set 
// and doing a mysql_fetch_assoc 
// for example, $sql = your query here 
// mysql_fetch_assoc($result); etc 

echo json_encode($arr); 
?> 

{"a":1,"b":2,"c":3,"d":4,"e":5} 

然後,你可以這樣做

<script type="text/javsacript"> 
var abc = "<? echo json_encode($arr);?>"; 
</script> 

OR

echo '<script type="text/javsacript"> 
     var abc ="'.json_encode($arr).'"; 
    </script>'; 
0

其實這是一個非常模糊的問題,但我認爲AJAX就是你要找 什麼。

編輯:當然JSON將試訓過,甚至可能會更直接...

+0

你不一定非要異步調用希望從數據庫獲取信息到JavaScript! – nickf 2009-09-22 07:34:32

+0

Mkkay,也許AJAX是一個輪子上的一個飛... – KB22 2009-09-22 07:36:19

1

// first, build your query: 
$sql = "SELECT name, email FROM users"; 

$result = mysql_query($sql); 

// then build up your data 
$rows = array(); 

while ($row = mysql_fetch_assoc($result)) { 
    $rows[] = $row; 
} 

//then write it in a way Javascript can understand: 

echo "<script type=\"text/javascript\">\n" 
    . "var users = " . json_encode($rows) . ";\n" 
    . "</script>"; 
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