[英]How pass a parameter to jQuery function
在這里,我使用此代碼使用JSON
創建HTML
表,在我的Web程序中,我使用PHP
生成了JSON
,現在我需要將JSON
作為參數傳遞給上述函數,我該怎么做。
<script type="text/javascript">
var data = [
{
"UserID": 1,
"UserName": "rooter",
"Password": "12345",
"Country": "UK",
"Email": "sac@gmail.com"
},
{
"UserID": 2,
"UserName": "binu",
"Password": "123",
"Country": "uk",
"Email": "Binu@gmail.com"
},
{
"UserID": 3,
"UserName": "cal",
"Password": "123",
"Country": "uk",
"Email": "cal@gmail.com"
},
{
"UserID": 4,
"UserName": "nera",
"Password": "1234",
"Country": "uk",
"Email": "nera@gmail.com"
},
{
"UserID": 4,
"UserName": "nera",
"Password": "1234",
"Country": "uk",
"Email": "nera@gmail.com"
}
];
$(document).ready(function () {
var html = '<table class="table table-striped">';
html += '<tr>';
var flag = 0;
$.each(data[0], function(index, value){
html += '<th>'+index+'</th>';
});
html += '</tr>';
$.each(data, function(index, value){
html += '<tr>';
$.each(value, function(index2, value2){
html += '<td>'+value2+'</td>';
});
html += '<tr>';
});
html += '</table>';
$('body').html(html);
});
</script>
我可以使用PHP
將JSON
值作為參數傳遞給此函數嗎
您可以使用php標簽作為參數傳遞
<?php
$data = '[
{
"UserID": 1,
"UserName": "rooter",
"Password": "12345",
"Country": "UK",
"Email": "sac@gmail.com"
},
{
"UserID": 2,
"UserName": "binu",
"Password": "123",
"Country": "uk",
"Email": "Binu@gmail.com"
},
{
"UserID": 3,
"UserName": "cal",
"Password": "123",
"Country": "uk",
"Email": "cal@gmail.com"
},
{
"UserID": 4,
"UserName": "nera",
"Password": "1234",
"Country": "uk",
"Email": "nera@gmail.com"
},
{
"UserID": 4,
"UserName": "nera",
"Password": "1234",
"Country": "uk",
"Email": "nera@gmail.com"
}
]';
?>
<script>
$(document).ready(function () {
var html = '<table class="table table-striped">';
html += '<tr>';
var flag = 0;
var data2 = <?php echo $data; ?>;
$.each(data2[0], function(index, value){
html += '<th>'+index+'</th>';
});
html += '</tr>';
$.each(data2, function(index, value){
html += '<tr>';
$.each(value, function(index2, value2){
html += '<td>'+value2+'</td>';
});
html += '<tr>';
});
html += '</table>';
$('body').html(html);
console.log(html);
});
</script>
我建議只在php中創建表,以便在加載時在頁面上,這對於seo和用戶體驗會更好。
但這是從php加載頁面時無需發出ajax請求即可打印javascript數據的方式。
關鍵是<?= json_encode(array); ?>
<?= json_encode(array); ?>
json編碼會將php數組/關聯數組解析為json對象。
我在單個腳本標簽中進行了此操作,以防解析或您的編輯器語法出錯。
<?php
$data = [
[ "UserID" => 1, "UserName" => "rooter", "Password" => "12345", "Country" => "UK", "Email" => "sac@gmail.com" ],
[ "UserID" => 2, "UserName" => "binu", "Password" => "123", "Country" => "uk", "Email" => "Binu@gmail.com" ],
];
?>
<script>data = <?= json_encode($data, JSON_NUMERIC_CHECK); ?></script>
$(document).ready(function () {
var html = '<table class="table table-striped">';
html += '<tr>';
var flag = 0;
$.each(data[0], function(index, value){
html += '<th>'+index+'</th>';
});
html += '</tr>';
$.each(data, function(index, value){
html += '<tr>';
$.each(value, function(index2, value2){
html += '<td>'+value2+'</td>';
});
html += '<tr>';
});
html += '</table>';
$('body').html(html);
});
</script>
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