duandu1915 2016-10-26 22:14 采纳率: 100%
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将PHP数组传递给JSON和JS:Uncaught SyntaxError:意外的标识符

i always get "Uncaught SyntaxError: Unexpected identifier" when i try to echo a PHP Variable to a JS Variable.

<script type="text/javascript">
    $(function () {
        var options = {
           float: false,
           removable: '.trash',
           removeTimeout: 100,
           acceptWidgets: '.grid-stack-item',
           resizable: { handles: 'e, se, s, sw, w' }
           };
        $('#grid').gridstack(options);

        var data = "<?php echo json_encode($serializedData); ?>";

My PHP:

$serializedData = array();
$str = "SELECT gri_id as 'id', gri_plugin as 'plugin', gri_gridContent as 'content', gri_gridPosX as 'x', gri_gridPosY as 'y', gri_gridSizeX as 'width', gri_gridSizeY as 'height' FROM tGrid WHERE gri_location = '$loc'";
$stmt = $db->prepare($str);
$stmt->execute();
$result = $stmt->fetchAll(PDO::FETCH_ASSOC);
foreach ($result as $value) {
  $function = $value["plugin"];
  $data = $function($value['id'], $db);
  $value['content'] = $data;
  array_push($serializedData, $value);
}
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1条回答 默认 最新

  • dpkt31779 2016-10-26 22:28
    关注

    Assuming JSON is a valid snippet of javascript, why should we wrap it into quotes? You expect data will become Object, but it becomes String for now.

    Do:

    var data = <?php echo json_encode($serializedData) ?>;
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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