phpmyadmin 在PHP中存储JOIN值并将其显示为表[已关闭]

eqoofvh9 于 2022-11-09 发布在 PHP

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我有以下SQL查询：

SELECT tbltasks.fldTaskNr
    ,tbltasks.fldRITMNr
    ,tbltasks.fldCHGNr
    ,tblci.fldCI
    ,tblgxp.fldGxP
    ,tblrequester.fldRequester
    ,tblstatus.fldStatus
    ,tbltasks.fldDescription
    ,tblresponsible.fldResponsible
    ,tbllocation.fldLocation
FROM tbltasks
LEFT JOIN tblCI ON tblci.pkCI = tbltasks.fkCI
LEFT JOIN tblgxp ON tblgxp.pkGxP = tbltasks.fkGxP
LEFT JOIN tblrequester ON tblrequester.pkRequester = tbltasks.fkRequester
LEFT JOIN tblstatus ON tblstatus.pkStatus = tbltasks.fkStatus
LEFT JOIN tblresponsible ON tblresponsible.pkResponsible = tbltasks.fkResponsible
LEFT JOIN tbllocation ON tbllocation.pkLocation = tbltasks.fkLocation

输出：x1c 0d1x
现在我想存储这个查询的值，并将它们输出为一个表，看起来应该像这样：

我得到了下面的表作为HTML，输出应该显示在这个表中，并重复本身，因为将有几个条目在数据库中：

<th>Task Nr.</th>
                    <th>RITM Nr.</th>
                    <th>CHG Nr.</th>
                    <th>CI</th>
                    <th>GxP</th>
                    <th>Task Requester</th>
                    <th>Task Status</th>
                    <th>Description</th>
                    <th>Responsible</th>
                    <th style="width: 201px;">Location</th>
                </tr>

我怎么能这样做？任何帮助都是感激不尽的！

phpmyadmin

来源：https://stackoverflow.com/questions/70462351/storing-join-values-in-php-and-showing-them-as-a-table

2条答案

按热度按时间

clj7thdc1#

这很简单，但过程有点长。首先确保为查询列指定一个名称值，如下所示：

SELECT tbltasks.fldTaskNr AS "TASK_NR"
,tbltasks.fldRITMNr AS "RITMNR"
,tbltasks.fldCHGNr AS "CHGNR"
,tblci.fldCI AS "FLDCI"
,tblgxp.fldGxP AS "GXP"
,tblrequester.fldRequester AS "REQUESTER"
,tblstatus.fldStatus AS "FLD_STATUS"
,tbltasks.fldDescription AS "FLD_DESCRIPTION"
,tblresponsible.fldResponsible AS "RESPONSIBLE"
,tbllocation.fldLocation AS "LOCATION" FROM tbltasks .. (rest of your joints)

然后使用此代码将查询转换为对象的数组

$data = array();
while ($row = mysqli_fetch_assoc($result)) {
    $data[] = $row; 
} 
return $data;

完成这一步后，我们就可以转到HTMLTable部件了

<tr>
<th>Task Nr.</th>
<th>RITM Nr.</th>
<th>CHG Nr.</th>
<th>CI</th>
<th>GxP</th>
<th>Task Requester</th>
<th>Task Status</th>
<th>Description</th>
<th>Responsible</th>
<th style="width: 201px;">Location</th>
</tr>

<?php
for($i = 0 ;$i < count($data); $i++){
echo '<tr>';
echo '<td>'.$data['TASK_NR'].'</td>';
echo '<td>'.$data['RITMNR'].'</td>';
echo '<td>'.$data['CHGNR'].'</td>';
echo '<td>'.$data['FLDCI'].'</td>';
echo '<td>'.$data['GXP'].'</td>';
echo '<td>'.$data['REQUESTER'].'</td>';
echo '<td>'.$data['FLD_STATUS'].'</td>';
echo '<td>'.$data['FLD_DESSCRIPTION'].'</td>';
echo '<td>'.$data['RESPONSIBLE'].'</td>';
echo '<td style="width: 201px;">'.$data['LOCATION'].'</td>';
echo '</tr>';
}
?>

希望这将解决您的问题，并确保您的文件扩展名是“.php”

赞(0）回复(0）举报 2022-11-09

scyqe7ek2#

您可以使用foearch将提取到html中的数据打印出来。

foreach ($data as $key => $val) {
    <tr>
       <td>
          <?php echo $val ?>
       </td>
    </tr>

}
您可以通过将上述代码应用于自己的表和数据来获得所需的结果。

赞(0）回复(0）举报 2022-11-09

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phpmyadmin 在PHP中存储JOIN值并将其显示为表[已关闭]

2条答案

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