有许多人在使用Ajax进行DataTables和服务器端处理时遇到以下错误，但是我已经尝试了很多人并且阅读了所有内容，没有任何成功。
DataTables警告（表格ID =＆＃39;示例＆＃39;） - 无效的JSON响应
我还检查了dev答案，他们说每次出现这个错误都是因为你有json格式问题，或数据库连接问题或列选择问题。 好吧，在我看来，我没有。
我的代码（来自这个例子：here）是（try.php）：
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<!-- DataTables CSS -->
<link rel="stylesheet" type="text/css" href="//cdn.datatables.net/1.10.9/css/jquery.dataTables.css">
<script type="text/javascript" charset="utf8" src="http://ajax.aspnetcdn.com/ajax/jQuery/jquery-1.8.2.min.js"></script>
<title>test</title>
</head>
<body>
<table id="example">
<thead>
<tr>
<th>ID</th>
<th>country Name</th>
<th>country Currency</th>
</tr>
</thead>
<tbody>
</tbody>
</table>
<script type="text/javascript" charset="utf8" src="http://ajax.aspnetcdn.com/ajax/jquery.dataTables/1.9.4/jquery.dataTables.min.js"></script>
<script>
$(document).ready(function() {
$('#example').dataTable( {
"bProcessing": true,
"bServerSide": true,
"sAjaxSource": "server_processing2.php"
} );
} );
</script>
</body>
</html>
这是server_processing2.php：
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Processing</title>
</head>
<body>
<?php
/*
* Script: DataTables server-side script for PHP and MySQL
* Copyright: 2010 - Allan Jardine
* License: GPL v2 or BSD (3-point)
*/
/* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
* Easy set variables
*/
/* Array of database columns which should be read and sent back to DataTables. Use a space where
* you want to insert a non-database field (for example a counter or static image)
*/
$aColumns = array('ID', 'country_Name', 'country_Currency' );
/* Indexed column (used for fast and accurate table cardinality) */
$sIndexColumn = "ID";
/* DB table to use */
$sTable = "CountryInfos";
/* Database connection information */
$gaSql['user'] = "...";
$gaSql['password'] = "...";
$gaSql['db'] = "...";
$gaSql['server'] = "...";
/* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
* If you just want to use the basic configuration for DataTables with PHP server-side, there is
* no need to edit below this line
*/
/*
* MySQL connection
*/
$gaSql['link'] = mysql_pconnect( $gaSql['server'], $gaSql['user'], $gaSql['password'] ) or
die( 'Could not open connection to server' );
mysql_select_db( $gaSql['db'], $gaSql['link'] ) or
die( 'Could not select database '. $gaSql['db'] );
/*
* Paging
*/
$sLimit = "";
if ( isset( $_GET['iDisplayStart'] ) && $_GET['iDisplayLength'] != '-1' )
{
$sLimit = "LIMIT ".mysql_real_escape_string( $_GET['iDisplayStart'] ).", ".
mysql_real_escape_string( $_GET['iDisplayLength'] );
}
/*
* Ordering
*/
if ( isset( $_GET['iSortCol_0'] ) )
{
$sOrder = "ORDER BY ";
for ( $i=0 ; $i<intval( $_GET['iSortingCols'] ) ; $i++ )
{
if ( $_GET[ 'bSortable_'.intval($_GET['iSortCol_'.$i]) ] == "true" )
{
$sOrder .= $aColumns[ intval( $_GET['iSortCol_'.$i] ) ]."
".mysql_real_escape_string( $_GET['sSortDir_'.$i] ) .", ";
}
}
$sOrder = substr_replace( $sOrder, "", -2 );
if ( $sOrder == "ORDER BY" )
{
$sOrder = "";
}
}
/*
* Filtering
* NOTE this does not match the built-in DataTables filtering which does it
* word by word on any field. It's possible to do here, but concerned about efficiency
* on very large tables, and MySQL's regex functionality is very limited
*/
$sWhere = "";
if ( $_GET['sSearch'] != "" )
{
$sWhere = "WHERE (";
for ( $i=0 ; $i<count($aColumns) ; $i++ )
{
$sWhere .= $aColumns[$i]." LIKE '%".mysql_real_escape_string( $_GET['sSearch'] )."%' OR ";
}
$sWhere = substr_replace( $sWhere, "", -3 );
$sWhere .= ')';
}
/* Individual column filtering */
for ( $i=0 ; $i<count($aColumns) ; $i++ )
{
if ( $_GET['bSearchable_'.$i] == "true" && $_GET['sSearch_'.$i] != '' )
{
if ( $sWhere == "" )
{
$sWhere = "WHERE ";
}
else
{
$sWhere .= " AND ";
}
$sWhere .= $aColumns[$i]." LIKE '%".mysql_real_escape_string($_GET['sSearch_'.$i])."%' ";
}
}
/*
* SQL queries
* Get data to display
*/
$sQuery = "
SELECT SQL_CALC_FOUND_ROWS ".str_replace(" , ", " ", implode(", ", $aColumns))."
FROM $sTable
$sWhere
$sOrder
$sLimit
";
$rResult = mysql_query( $sQuery, $gaSql['link'] ) or die(mysql_error());
/* Data set length after filtering */
$sQuery = "
SELECT FOUND_ROWS()
";
$rResultFilterTotal = mysql_query( $sQuery, $gaSql['link'] ) or die(mysql_error());
$aResultFilterTotal = mysql_fetch_array($rResultFilterTotal);
$iFilteredTotal = $aResultFilterTotal[0];
/* Total data set length */
$sQuery = "
SELECT COUNT(".$sIndexColumn.")
FROM $sTable
";
$rResultTotal = mysql_query( $sQuery, $gaSql['link'] ) or die(mysql_error());
$aResultTotal = mysql_fetch_array($rResultTotal);
$iTotal = $aResultTotal[0];
/*
* Output
*/
$output = array(
"sEcho" => intval($_GET['sEcho']),
"iTotalRecords" => $iTotal,
"iTotalDisplayRecords" => $iFilteredTotal,
"aaData" => array()
);
while ( $aRow = mysql_fetch_array( $rResult ) )
{
$row = array();
for ( $i=0 ; $i<count($aColumns) ; $i++ )
{
if ( $aColumns[$i] == "version" )
{
/* Special output formatting for 'version' column */
$row[] = ($aRow[ $aColumns[$i] ]=="0") ? '-' : $aRow[ $aColumns[$i] ];
}
else if ( $aColumns[$i] != ' ' )
{
/* General output */
$row[] = $aRow[ $aColumns[$i] ];
}
}
$output['aaData'][] = $row;
}
echo json_encode( $output );
?>
</body>
</html>
我做过的任何尝试都给我发了错误，但这是我从server_processing2.php收到的ARRAY：
{"sEcho":0,
"iTotalRecords":"287",
"iTotalDisplayRecords":"287",
"aaData":[
["10768","Western Sahara","EH"],["10767","Vietnam","VND"],
["10765","Uzbekistan","UZS"],["10766","Venezuela","VEF"],
["10764","USA","USD"],["10763","Uruguay","UYU"],["10762","United Kingdom","GBP"],
["10761","United Arab Emirates","AED"],["10760","Ukraine","UAH"],
["10759","Uganda","UGX"],["10758","Turkmenistan","TMT"],.........
["10546","Argentina","ARS"],["10545","Albania","ALL"],
["10544","Afghanistan","AFN"]
]
}
所以我可以假设：
那为什么它不起作用？
更新
有效。 （我猜）有两个问题：
答案 0 :(得分：1)
您的server_processing2.php中包含HTML。删除<?php和?>标记之间除PHP代码之外的所有内容，以便仅返回JSON。