PHP Ajax回顯HTML表
[英]PHP Ajax echo HTML table
問題描述
將表格中的一些數據從PHP輸出到HTML時出現問題。 當我點擊Submit時，輸出沒有真正格式化為HTML表。 代碼已更新，因此現在一切正常。
<!DOCTYPE HTML PUBLIC
"-//W3C//DTD HTML 4.01 Transitional//EN"
"http://www.w3.org/TR/html401/loose.dtd">
<html>
<head>
<title>AJAX Database</title>
</head>
<body>
Type in the location <input type="text" id="name">
<input type="submit" id="name-submit" value="Check Inventory">
<div id="name-data"></div>
<script src="http://code.jquery.com/jquery-1.8.0.min.js"></script>
<script src="js/global.js"></script>
</body>
</html>
<div id="name-data"></div>就像是單擊“提交”按鈕后所有輸出的占位符。
這是PHP代碼。 我試圖在PHP代碼之外添加HTML標記，但無法正常工作。
<?php
// '.post' could be '.get' here but global.js has to be the same, .post is faster
if(isset($_POST['name']) === true && empty($_POST['name']) === false) {
require '../db/connect.php';
// id, location, quantity, count_sheet, remark
$query = mysql_query("
SELECT `sheet0_100`.`id`, `sheet0_100`.`location`,
`sheet0_100`.`quantity`, `sheet0_100`.`count_sheet`, `sheet0_100`.`remark`
FROM `sheet0_100`
WHERE `sheet0_100`.`location` = '" . mysql_real_escape_string(trim($_POST['name'])) . "'
");
// echo (mysql_num_rows($query) !== 0) ? mysql_fetch_row($query) : 'Location not found..';
if(mysql_num_rows($query) !== 0) {
// Find out how many rows are available
$rowsFound = @ mysql_num_rows($query);
// Report how many rows were found
echo "{$rowsFound} records found matching your criteria. \n";
echo "There are $rowsFound items in this location.\n" . "\xA";
echo "\n<table>";
// Display a table contains all the items
while ($rowsFound = mysql_fetch_row($query)) {
// ... start a TABLE row ...
echo "\n<tr>";
// ... and print out each of the attributes in that row as a
// separate TD (Table Data).
foreach($rowsFound as $data)
echo "\n\t<td> {$data} </td>";
// Finish the row
echo "\n</tr>";
}
// Then, finish the table
echo "</table>";
}
else {
echo "Location not found..";
}
}
?>
我的JS文件，
$('input#name-submit').on('click', function() {
var name = $('input#name').val();
if ($.trim(name) != '') {
// '.post' could be '.get' here but name.php has to be the same, .post is faster
$.post('ajax/name.php', {name: name}, function(data) {
// $('div#name-data').text(data);
// jQuery('#name-data').html(RESPONSE);
$('div#name-data').html(data);
});
}
});
2 個解決方案
解決方案1
3 已采納 2014-10-02 20:55:31
使用jQuery('#name-data').html(RESPONSE);
請參閱： http ： //api.jquery.com/html/
解決方案2
-1 2014-10-02 21:01:45
如果要使用php，則應將文本存儲在這樣的變量中：
if(mysql_num_rows($query) !== 0) {
// Find out how many rows are available
$rowsFound = @ mysql_num_rows($query);
// Report how many rows were found
echo "{$rowsFound} records found matching your criteria. \n";
echo "There are $rowsFound items in this location.\n" . "\xA";
$table = "<table>";
// Display a table contains all the items
while ($rowsFound = mysql_fetch_row($query)) {
// ... start a TABLE row ...
$table .="<tr>";
// ... and print out each of the attributes in that row as a
// separate TD (Table Data).
foreach($rowsFound as $data)
$table .= "<td> {$data} </td>";
// Finish the row
$table .= "</tr>";
}
// Then, finish the table
$table .= "</table>";
echo $table;
}
else {
echo "Location not found..";
}
將PHP轉換為HTML表
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