对此帖与我之前的帖子之间的相似之处表示道歉。我很感激，如果有人能再次帮助我找到我出错的地方。其他一切似乎都运转正常，但令我困惑的是对AJAX成功功能的“安静”反应。控制台中没有任何东西。
我已经使用json_encode和print_r函数测试了JSON输出并获得了以下内容 - 所以我假设JSON字符串应该可以使用AJAX：
Array
(
[proj_start_date] => 2017-04-17
[proj_end_date] => 2018-04-30
[wo_nbr_new] => 10002-06
)
{"proj_start_date":"2017-04-17","proj_end_date":"2018-04-30","wo_nbr_new":"10002-06"}
以下是主文件的代码：
<?php
include 'connect_db.php';
$sql = "SELECT * FROM projects ORDER BY proj_nbr";
$result = mysqli_query($connect,$sql);
$rowCount = mysqli_num_rows($result);
?>
<!DOCTYPE html>
<html lang="en">
<head>
<title>Add New Work Order</title>
<script
src="https://code.jquery.com/jquery-3.2.1.min.js"integrity="sha256-hwg4gsxgFZhOsEEamdOYGBf13FyQuiTwlAQgxVSNgt4="
crossorigin="anonymous">
</script>
<link rel="stylesheet" type="text/css" href="styles.css">
</head>
<body>
<h1>Add New Work Order</h1>
<div id="forms-add" name="forms-add">
<form action="add_workorders.php" method="POST">
<label>Work Order Number (Auto-generated):</label>
<input type="text" id="wo_nbr" name="wo_nbr"size="8"maxlength="8" value = "" readonly style="float: right">
<br><br>
<fieldset>
<legend>Project Details</legend>
<label>Project Number:</label>
<select class= "selects" id="proj_nbr" name="proj_nbr" required onchange="">
<option value="">Select a project </option>
<?php
if($rowCount>0)
{
while ($row=mysqli_fetch_assoc($result))
{
echo '<option value="'.
$row['proj_id'].'">'.
$row['proj_nbr'].
' - '.
$row['proj_desc'].
' </option>';
}
}
else
{
echo '<option value="">Project not available</option>';
}
?>
</select>
<br><br>
<label>Start Date:</label>
<input type="text" id="proj_start_date" name="proj_start_date"size="8"maxlength="8" value = "" readonly style="float: right">
<br><br>
<label>End Date:</label>
<input type="text" id="proj_end_date" name="proj_end_date"size="8"maxlength="8" value = "" readonly style="float: right">
</fieldset>
<br><br>
<button type="submit" name="submit" >Save Work Order</button>
</form>
</div>
</body>
<script type="text/javascript" src="test_ajax.js"></script>
<script>
$(document).ready(function()
{
$('#proj_nbr').change(function()
{
var id=$('#proj_nbr').val();
//alert(id); //this works ok
if (id != '')
{
$.ajax({
url: "get_proj_nbrs2.php",
method:"POST",
data: {id:id}, //data to SEND to PHP file
dataType: "JSON",
success: function(output)
{
console.log(output); //this doesn't return anything in the console??
$('#wo_nbr').val(output.wo_nbr_new);
$('#proj_start_date').val(output.proj_start_date);
$('#proj_end_date').val(output.proj_end_date);
}
});
}
else
{
alert("Please select a Project");
}
});
});
</script>
</html>
以下是PHP文件中的代码：
<?php
include 'connect_db.php';
if (isset($_POST['id']) && !empty($_POST['id']))
{
$sql2 = "SELECT p.proj_nbr as wo_proj_nbr,p.start_date as proj_start_date,p.end_date as proj_end_date, MAX(w.wo_nbr) AS wo_nbr,
CASE WHEN SUBSTRING(MAX(w.wo_nbr),7,2)+1 <= 9 THEN CONCAT(p.proj_nbr,'-0',SUBSTRING(MAX(w.wo_nbr),7,2)+1)
ELSE CONCAT(p.proj_nbr,'-',SUBSTRING(MAX(w.wo_nbr),7,2)+1) END AS wo_nbr_new
FROM workorders as w
INNER JOIN projects as p on p.proj_id = w.proj_id
WHERE w.proj_id = '".$_POST['id']."'";
$result2 = mysqli_query($connect,$sql2);
while($row=mysqli_fetch_array($result2))
{
if($result2 ==true)
{
$proj_nbr = $row['wo_proj_nbr'];
$output['proj_start_date'] = $row['proj_start_date'];
$output['proj_end_date'] = $row['proj_end_date'];
if ($row['wo_nbr_new'] != NULL)
{
$output['wo_nbr_new'] = $row['wo_nbr_new'];
echo json_encode($output);
}
elseif($row['wo_nbr_new'] == NULL)
{
$output['wo_nbr_new'] = $proj_nbr."-01";
echo json_encode($output);
}
}
}
}?>
答案 0 :(得分：0)
请确认您的id值不为空且与数据库字段值匹配。如果是这样，那么改变“id”＃39;到了＆＃39; theid＆＃39; （可能是保留字）在ajax例如
中 data: {theid:id}, //data to SEND to PHP file
并在php中。
if (isset($_POST['theid']) && !empty($_POST['theid'])) {
$sql2 = "SELECT p.proj_nbr as wo_proj_nbr,p.start_date as proj_start_date,p.end_date as proj_end_date, MAX(w.wo_nbr) AS wo_nbr,
CASE WHEN SUBSTRING(MAX(w.wo_nbr),7,2)+1 <= 9 THEN CONCAT(p.proj_nbr,'-0',SUBSTRING(MAX(w.wo_nbr),7,2)+1)
ELSE CONCAT(p.proj_nbr,'-',SUBSTRING(MAX(w.wo_nbr),7,2)+1) END AS wo_nbr_new
FROM workorders as w
INNER JOIN projects as p on p.proj_id = w.proj_id
WHERE w.proj_id = '".$_POST['theid']."'";
// code goes here
}
答案 1 :(得分：0)
<?php
include 'connect_db.php';
if (isset($_POST['id']) && !empty($_POST['id']))
{
$result = array();
$sql2 = "SELECT p.proj_nbr as wo_proj_nbr,p.start_date as proj_start_date,p.end_date as proj_end_date, MAX(w.wo_nbr) AS wo_nbr,
CASE WHEN SUBSTRING(MAX(w.wo_nbr),7,2)+1 <= 9 THEN CONCAT(p.proj_nbr,'-0',SUBSTRING(MAX(w.wo_nbr),7,2)+1)
ELSE CONCAT(p.proj_nbr,'-',SUBSTRING(MAX(w.wo_nbr),7,2)+1) END AS wo_nbr_new
FROM workorders as w
INNER JOIN projects as p on p.proj_id = w.proj_id
WHERE w.proj_id = '".$_POST['id']."'";
$result2 = mysqli_query($connect,$sql2);
if($result->num_rows > 0)
{
while($row=mysqli_fetch_array($result2))
{
$proj_nbr = $row['wo_proj_nbr'];
$output['proj_start_date'] = $row['proj_start_date'];
$output['proj_end_date'] = $row['proj_end_date'];
if ($row['wo_nbr_new'] != NULL)
{
$output['wo_nbr_new'] = $row['wo_nbr_new'];
}
elseif($row['wo_nbr_new'] == NULL)
{
$output['wo_nbr_new'] = $proj_nbr."-01";
}
array_push($result,$output);
}
echo json_encode($result);
}
else
{
echo "no rows found";
}
}?>
你回应每一行，所以你将从ajax中获得成功函数的结果作为每个单独的刺痛。所以它无法解析。希望这种方法能解决你的问题。
答案 2 :(得分：0)
在方法中使用type。方法POST依赖于jquery版本
$.ajax({
url: "get_proj_nbrs2.php",
type:"POST",
data: {id:id}, //data to SEND to PHP file
dataType: "JSON",
success: function(output)
{
}
});