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我创建了一个报表,其中包含2个下拉列表和上表中的表格。我想过滤数据并使用下拉列表选项生成报告。第一个下拉列表包含来自数据库的所有表名,第二个下拉列表具有“项目”列,每个表具有相同的项目值。就像我们从表3和项目选项1做出报告一样。我们应该做出报告。从php和mysql数据库中获取数据(创建报告)
这里是我的HTML代码:
<div class="row">
<div class="col-lg-12">
<div class="ibox float-e-margins">
<div class="ibox-title">
<h5>Report by Items</h5>
</div>
<div class="ibox-content">
<form method="post" class="form-horizontal">
<div class="form-group"><label class="col-sm-4 control-label"Stocks</label>
<div class="col-sm-4"><select class="form-control m-b" name="Stock">
<option></option>
<option>Stock</option>
<option>Stock1</option>
<option>Stock2</option>
</select>
</div>
</div>
<div class="form-group"><label class="col-sm-4 control-label">Items</label>
<div class="col-sm-4">
<select class="form-control m-b" name="Items">
<option></option>
<option>Item1</option>
<option>Item2</option>
<option>Item3</option>
<option>Item4</option>
</select>
</div>
</div>
</div>
<div class="col-lg-25">
<div class="ibox float-e-margins">
</div>
</div>
<div class="form-group">
<div class="col-sm-4 col-sm-offset-5">
<button class="btn btn-white" type="submit">Cancel</button>
<button class="btn btn-primary" type="submit" name=submit>Run Report</button>
</div>
</div>
</form>
</div>
</div>
</div>
</div>
</div>
PHP和表的代码,我尝试是:
<div class="ibox-content">
<table class="table table-striped table-bordered table-hover dataTables-example" >
<thead>
<tr>
<th>No</th>
<th>Quantity</th>
<th>Date</th>
<th>Sold</th>
<th>Total</th>
</tr>
</thead>
<tbody>
<?php
$mysqli = new mysqli('localhost', 'user', 'pass', 'mis_db');
if (mysqli_connect_error()) {
echo mysqli_connect_error();
exit();
}
if($_POST['Stock']=='Stock1')
{
$stck1 = 'Stock1';
}
if($_POST['Stock']=='Stock2')
{
$stck1 = 'Stock2';
}
if($_POST['Stock']=='Stock1')
{
$stck1 = 'Stock3';
}
if($_POST['Items']=='Item1')
{
$itm = 'Item1';
}
if($_POST['Items']=='Item2')
{
$itm = 'Item2';
}
if($_POST['Items']=='Item3')
{
$itm = 'Item3';
}
if (isset($_POST['submit'])) {
$query = 'SELECT * FROM .$stck1 where Items=.$itm';
$data = mysqli_query($mysqli, $query) ;
if (!$data) {
echo("Error description: " . mysqli_error($mysqli));
} else {
while ($row = mysqli_fetch_array($data)) {
echo "<tr>
<td>" . $row['No'] . "</td>
<td>" . $row['Qty'] . "</td>
<td>" . $row['date'] . "</td>
<td>" . $row['Sold'] . "</td>
<td>" . $row['Total'] . "</td>
</tr>";
}
}
}
?>
</tbody>
<tfoot>
</tfoot>
</table>
</div>
我怀疑MySQL查询。
它给出了这样的错误:错误描述:在“where子句中” –
是什么,你的数据库表的样子未知列“物品”? – pmahomme
对不起,我没有?喜欢? –