gpt4 book ai didi

php - 每个结果都出现在表格而不是行中

转载 作者:行者123 更新时间:2023-11-28 11:01:51 25 4
gpt4 key购买 nike

这是我用于搜索的 php 代码,但为什么结果不像在带有行的单个表中那样出现。我尝试像 '<td>','<div>' . $row['$accountcode']. '</div>','</td>' 这样更改 PHP但是有一个错误。我不擅长编码,所以我不确定我该怎么做。所以这是它的样子:

like this .

所有结果都显示在表格中。这是因为我的 CSS 而发生的吗?

table,tr,td,th {
border: 1px solid black;
white-space: normal;
}

table tr td:nth-child(1) {
background-color: red;
}

table {
/*table-layout: fixed;
width: 140%;*/
margin-left: 190px;
margin-right: 10px;
display: block;
overflow-y: auto;
height: 450px;
}

th > td > div {
max-width: 300px;
min-width: 50px;
word-wrap: break-word;
overflow-wrap: break-word;
}
if(isset($_POST['searchq']) && $_POST['searchq'] != "")
{
if(isset($_POST['searchopt']) && $_POST['searchopt'] != "")
{
$escsearch = mysqli_real_escape_string($conn, $_POST['searchq']);
$searchval = preg_replace('#[^a-z 0-9]#i', '', $escsearch);
$opt = $_POST['searchopt'];

switch($opt)
{
case "dpt":
$query = "SELECT * FROM coi_system where department LIKE '%$searchval%'";
$res = mysqli_query($conn, $query) or die(mysqli_error());
$count = mysqli_num_rows($res);
if($count > 0)
{
$output = "$count results for <strong>$escsearch</strong>.";

while($row = mysqli_fetch_array($res))
{
$accountcode = $row['accountcode'];
$department = $row['department'];
$person_in_charge = $row['person_in_charge'];
$project_title = $row['project_title'];
$objective = $row['objective'];
$how_to_do = $row['how_to_do'];
$activities = $row['activities'];
$project_started = $row['project_started'];
$project_completed = $row['project_completed'];
$target_cost_saving = $row['target_cost_saving'];
$costsaving_afterjustification = $row['costsaving_afterjustification'];
$costsaving_monthly = $row['costsaving_monthly'];

echo "<table>
<tr>
<td><div>Account Code</div></td>
<td><div>Department</div></td>
<td><div>Person in charge</div></td>
<td><div>Project title</div></td>
<td><div>Objective</div></td>
<td><div>How To Do</div></td>
<td><div>Activities</div></td>
<td><div>Project Started</div></td>
<td><div>Project Completed</div></td>
<td><div>Target Cost Saving</div></td>
<td><div>Cost Saving After Justification</div></td>
<td><div>Cost Saving Monthly</div></td>
</tr>
<tr>
<td><div>$accountcode</div></td>
<td><div>$department</div></td>
<td><div>$person_in_charge</div></td>
<td><div>$project_title</div></td>
<td><div>$objective</div></td>
<td><div>$how_to_do</div></td>
<td><div>$activities</div></td>
<td><div>$project_started</div></td>
<td><div>$project_completed</div></td>
<td><div>$target_cost_saving</div></td>
<td><div>$costsaving_afterjustification</div></td>
<td><div>$costsaving_monthly</div></td>
</tr>
</table>"; }
}else{
$output = "<p>No records found.</p>";
}
break;
}
}
}

?

最佳答案

它没有出现在单个表中,因为您为每条记录创建了一个新表。只需取出 <table来自 while 循环的标记。

echo"<table>";
while($row = mysqli_fetch_array($res))
{
$accountcode = $row['accountcode'];
$department = $row['department'];
$person_in_charge = $row['person_in_charge'];
$project_title = $row['project_title'];
$objective = $row['objective'];
$how_to_do = $row['how_to_do'];
$activities = $row['activities'];
$project_started = $row['project_started'];
$project_completed = $row['project_completed'];
$target_cost_saving = $row['target_cost_saving'];
$costsaving_afterjustification = $row['costsaving_afterjustification'];
$costsaving_monthly = $row['costsaving_monthly'];

echo "
<tr>
<td><div>Account Code</div></td>
<td><div>Department</div></td>
<td><div>Person in charge</div></td>
<td><div>Project title</div></td>
<td><div>Objective</div></td>
<td><div>How To Do</div></td>
<td><div>Activities</div></td>
<td><div>Project Started</div></td>
<td><div>Project Completed</div></td>
<td><div>Target Cost Saving</div></td>
<td><div>Cost Saving After Justification</div></td>
<td><div>Cost Saving Monthly</div></td>
</tr>
<tr>
<td><div>$accountcode</div></td>
<td><div>$department</div></td>
<td><div>$person_in_charge</div></td>
<td><div>$project_title</div></td>
<td><div>$objective</div></td>
<td><div>$how_to_do</div></td>
<td><div>$activities</div></td>
<td><div>$project_started</div></td>
<td><div>$project_completed</div></td>
<td><div>$target_cost_saving</div></td>
<td><div>$costsaving_afterjustification</div></td>
<td><div>$costsaving_monthly</div></td>
</tr>
"; }

echo " </table>";

关于php - 每个结果都出现在表格而不是行中,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48679626/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com