我试图通过ajax将7个jquery复选框的值发送到php。我试图将值放在一个数组中并在ajax中序列化该数组。最后，我想使用checked复选框的值作为条件是MySQL WHERE子句。当我尝试在php中使用结果时，我的代码失败，并且该页面中的表将不会显示。但是，如果我在该php页面中注释掉$ _POST行，则会显示我的表格。在浏览器中检查代码时，我会得到一个500（内部服务器错误）表，应该显示该表。我也有ajax显示错误结果如下：
[object Object]
代码：
我的HTML代码：
<label for="prestage_select">Prestage</label>
<input type="checkbox" name="revenue_checkboxes[]" id="prestage_select" value="Prestage">
<label for="validation_select">Validation</label>
<input type="checkbox" name="revenue_checkboxes[]" id="validation_select" value="Validation">
<label for="scheduling_select">Scheduling</label>
<input type="checkbox" name="revenue_checkboxes[]" id="scheduling_select" value="Scheduling">
<label for="production_select">Production</label>
<input type="checkbox" name="revenue_checkboxes[]" id="production_select" value="Production">
<label for="needsBOL_select">Needs BOL</label>
<input type="checkbox" name="revenue_checkboxes[]" id="needsBOL_select" value="Needs BOL">
<label for="shpAcct2Close_select">Shipped: Account to Close</label>
<input type="checkbox" name="revenue_checkboxes[]" id="shpAcct2Close_select" value="Shipped: Acctg. To Close Out">
<label for="movedToComplete_select">Moved to Complete for Selected Period</label>
<input type="checkbox" name="revenue_checkboxes[]" id="movedToComplete_select" value="Complete">
我的Ajax代码：
j("#create_submit").click(function(){
//Send Revenue Status Values to php using ajax.
j.ajax({
method: 'POST',
url: 'revenue_report.php',
data: j('[name="revenue_checkboxes[]"]').serialize(),
success: function( response ) {
j('#fieldset_ReportDiv').html(response);
}
});
//send Revenue Date values to php using ajax.
var revenuefrom = j('#revenuefrom').val();
var revenueto = j('#revenueto').val();
j.ajax ({
method: 'POST',
url: "revenue_report.php",
data: { revenuefromtext: revenuefrom, revenuetotext: revenueto },
success: function( response ) {
j('#fieldset_ReportDiv').html(response);
}
});
我的PHP代码：
<?php
include('inc.php');
//Get date range.
$revenuefromajax=$_POST['revenuefromtext'];
$revenuetoajax=$_POST['revenuetotext'];
$revenuefromstring = strtotime($revenuefromajax);
$revenuetostring = strtotime($revenuetoajax);
$revenuefrom=date("Y-m-d", $revenuefromstring);
$revenueto=date("Y-m-d", $revenuetostring);
//Get selected Status Values.
$revenuecheckboxes=$_POST('revenue_checkboxes'); //the page loads properly when this line is commented out. As soon as I remove the comment, it breaks.
//connect to the database
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if(mysqli_connect_errno() ) {
printf('Could not connect: ' . mysqli_connect_error());
exit();
}
//echo 'MySQL Connected successfully.'."<BR>";
$conn->select_db("some database name"); /////Database name has been changed for security reasons/////////
if(! $conn->select_db('some database name') ) {
echo 'Could not select database. '."<BR>";
}
// echo 'Successfully selected database. '."<BR>";
//Select Data and Display it in a table.
$sql = "SELECT invoices.id, invoices.orderdate, invoices.stagestatus, FORMAT(TRIM(LEADING '$' FROM invoices.totalprice), 2) AS totalprice, clients.company, lineitems.invoiceid, FORMAT((lineitems.width * lineitems.height) /144, 2 ) AS sqft, lineitems.quantity AS qty, FORMAT((invoices.totalprice / ((lineitems.width * lineitems.height) /144)), 2) as avgsqftrevenue, FORMAT((TRIM(LEADING '$' FROM invoices.totalprice) / lineitems.quantity), 2) AS avgunitrevenue
FROM clients
INNER JOIN invoices ON clients.id = invoices.clientid
INNER JOIN lineitems ON invoices.id = lineitems.invoiceid
WHERE invoices.orderdate BETWEEN '".$revenuefrom."' AND '".$revenueto."'
ORDER BY invoices.id DESC";
$result = $conn->query($sql);
echo "<table id='revenueReportA' align='center' class='report_DT'>
<tr>
<th>Customer</th>
<th>SG</th>
<th>Revenue</th>
<th>SQ FT</th>
<th>AVG Revenue Per SQ FT</th>
<th>Number of Units</th>
<th>AVG Revenue Per Unit</th>
</tr>";
if ($result = $conn->query($sql)) {
// fetch associative array
while ($row = $result->fetch_assoc()) {
echo "<tr>";
echo "<td>" . $row['company'] . "</td>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" ."$". $row['totalprice'] . "</td>";
echo "<td>" . $row['sqft'] ." ". "ft<sup>2</sup>". "</td>";
echo "<td>" ."$". $row['avgsqftrevenue'] . "</td>";
echo "<td>" . $row['qty'] . "</td>";
echo "<td>" ."$". $row['avgunitrevenue'] . "</td>";
echo "</tr>";
}
echo "</table>";
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//Free the result variable.
$result->free();
}
//Close the Database connection.
$conn->close();
?>
我已经尝试了几种不同的建议，将值发送到php但ajax仍然失败。
注意：我包含了另一个针对revenueto的ajax调用和来自日期的收入。此调用成功，只要复选框ajax的数据被注释掉，我的表就会根据这些日期正确显示。
注2：道歉，因为我是ajax的新手。
谢谢！
答案 0 :(得分：1)
$_POST不正确。序列化的是输入的名称/值。输入名称中的[]会自动被$_POST选为关键字revenue_checkboxes的数组
试
$revenuecheckboxes = $_POST['revenue_checkboxes'];// should be an array of the checked checkboxes
答案 1 :(得分：0)
您获取数组的方式 $revenuecheckboxes=$_POST('[name="revenue_checkboxes[]"]');无效（此处不要使用javascript方式; D）应该为$revenuecheckboxes=$_POST['revenue_checkboxes'];
php将自动发现它的数组，并将'revenuecheckboxes'初始化为数组及其值