mysql db result转换为json数组

Question

我正在尝试编写一个自动完成,当PHP页面加载时,自动完成项加载一次.使用从mysql DB获取的项目,我创建了一个像这样的json数组

<?php
$bnkArray = array();
$sql_bnk = mysql_query("SELECT BName, BCode, ID  FROM bank");
while($rBnk = mysql_fetch_array($sql_bnk)){

    $bnkDet = array(
        'label' => $rBnk['BName'],
        'value' => $rBnk['BName'],
        'otherDetails' => $rBnk['BName'].'||'. $rBnk['BCode'].'||'. $rBnk['ID'] 
    );  
    array_push($bnkArray, $bnkDet);
}

?>

我需要这个数组像这个javascript数组

<script>

var bankSource11 = [
      {
        value: "jquery",
        label: "jQuery",
        otherDetails: "the write less, do more, JavaScript library",

      },
      {
        value: "jquery-ui",
        label: "jQuery UI",
        otherDetails: "the official user interface library for jQuery",

      },
      {
        value: "sizzlejs",
        label: "Sizzle JS",
        otherDetails: "a pure-JavaScript CSS selector engine",

      }
    ];
</script>

如果我在我的汽车完成中这样调用这个数组,这是行不通的

var bankSource = [<?php echo  $bnkArray; ?>];

什么是这种数组类型..如何做到这一点.

这是自动完成部分

$(this).autocomplete({
            minLength: 0,
            source: bankSource,
            focus: function( event, ui ) {
                $(this).val( ui.item.label );
                return false;
            },
            select: function( event, ui ) {
                console.log(ui.item.value +' ____ ' + ui.item.otherDetails);
                $( "#project" ).val( ui.item.label );
                $( "#project-id" ).val( ui.item.value );
                $( "#project-description" ).html( ui.item.otherDetails );

                return false;
              }
            })


        }

Answer 1

请更换

var bankSource = [<?php echo  $bnkArray; ?>];

和

var bankSource = <?php echo json_encode($bnkArray); ?>;