我有一张考试表，其中有一个编辑部分。当我选择其中一项考试进行编辑时，它不会采用考试的 ID。我不知道为什么。这是我的代码:
这是显示表格的主要页面内容
<div class="view">
<table class="tabel">
<tr class="tepprand">
<td>Data</td>
<td>An</td>
<td>Materia</td>
<td>Profesor</td>
<td>Asistent</td>
<td>Sala</td>
<td>Tip Examen</td>
</tr>
<?php
$sqll = "SELECT r.id AS id , r.data AS data , ra.an AS an, rm.numemat AS numemat, rp1.numep AS numep1 , rp1.prenumep AS prenumep1, rp2.numep AS numep2, rp2.prenumep AS prenumep2, rs.salaa AS salaa , re.tip AS tip
FROM examen_programat AS r
LEFT JOIN an ra ON (r.id_ann=ra.id_an)
LEFT JOIN materii rm ON (r.id_mate=rm.id_mat)
LEFT JOIN profesor rp1 ON (r.id_prof1=rp1.id_prof)
LEFT JOIN profesor rp2 ON (r.id_prof2=rp2.id_prof)
LEFT JOIN sala rs ON (r.id_sala=rs.id_s)
LEFT JOIN examen re ON (r.id_tipp = re.id_tip)
ORDER by data";
$result = mysqli_query($link,$sqll);
while($rows=mysqli_fetch_array($result,MYSQLI_BOTH)) {
?>
<tr>
<td><?php echo $rows['data']; ?></td>
<td><?php echo $rows['an']; ?></td>
<td><?php echo $rows['numemat']; ?></td>
<td><?php echo $rows['numep1']." ".$rows['prenumep1']; ?></td>
<td><?php echo $rows['numep2']." ".$rows['prenumep2']; ?></td>
<td><?php echo $rows['salaa']; ?></td>
<td><?php echo $rows['tip']; ?></td>
<td><a href="editexamene.php?id=<? echo $rows['id']; ?>"> Edit </a></td>
<td><a href="deleteexamene.php?id=<? echo $rows['id']; ?>"> Stergere </a></td>
</tr>
<?php
}
?>
</table>
</div>
这是当我按下 EDIT 时发生的事情:
<div class="view">
<?php
$idul=$_GET['id'];
$str = "SELECT * FROM examen_programat WHERE id =$idul ";
$rez = mysqli_query($link,$str);
$row = mysqli_fetch_array($rez);
$an=$row['id_ann'];
$materie=$row['id_mate'];
$profesor=$row['id_prof1'];
$asistent=$row['id_prof2'];
$salaa=$row['id_sala'];
$tip=$row['id_tipp'];
?>
<form name="tabel" method="post" action="updateexamen.php">
<input type="hidden" name="idu" value="<?php echo " $idul" ?>">
<table>
<tr>
<td>Data</td>
<td><input type="date" name="data" value="<?php echo $row['data'] ?>" required="required"/><br></td>
</tr>
<tr>
<td>An</td>
<td>
<?php
$sql_year="SELECT * FROM an";
$rez_year = mysqli_query($link,$sql_year);
echo "<select name=\"year\" >";
while($year=mysqli_fetch_array($rez_year))
{
echo "
<option value=\"".$year['id_an'];
if ($year[
'id_an']==$an)
echo "\" selected=\"selected\">".$year['grupa']."</option>\n";
else
echo "\">".$year['grupa']."</option>\n";
}
echo "</select>";
?><br>
</td>
</tr>
<tr>
<td>Materie</td>
<td>
<?php
$sql_mat="SELECT * FROM materii";
$rez_mat = mysqli_query($link,$sql_mat);
echo "<select name=\"mat\" >";
while($mat=mysqli_fetch_array($rez_mat))
{
echo "
<option value=\"".$mat['id_mat'];
if ($mat[
'id_mat']==$materie)
echo "\" selected=\"selected\">".$mat['numemat']."</option>\n";
else
echo "\">".$mat['numemat']."</option>\n";
}
echo "</select>";
?><br>
</td>
</tr>
<tr>
<td>Profesor</td>
<td>
<?php
$sql_proff="SELECT * FROM profesor";
$rez_proff = mysqli_query($link,$sql_proff);
echo "<select name=\"proff\" >";
while($proff=mysqli_fetch_array($rez_proff))
{
echo "
<option value=\"".$proff['id_prof'] ;
if($proff[
'id_prof']==$profesor)
echo "\" selected=\"selected\">".$proff['numep']." ".$proff['prenumep']."</option>\n";
else
echo "\">".$proff['numep']." ".$proff['prenumep']."</option>\n";
}
echo "</select>";
?><br>
</td>
</tr>
<tr>
<td>Asistent</td>
<td>
<?php
$sql_profff="SELECT * FROM profesor";
$rez_profff = mysqli_query($link,$sql_profff);
echo "<select name=\"profff\" >";
while($profff=mysqli_fetch_array($rez_profff))
{
echo "
<option value=\"".$profff['id_prof'];
if($profff[
'id_prof']==$asistent)
echo "\" selected=\"selected\">".$profff['numep']." ".$profff['prenumep']."</option>\n";
else
echo "\">".$profff['numep']." ".$profff['prenumep']."</option>\n";
}
echo "</select>";
?><br>
</td>
</tr>
<tr>
<td>Sala</td>
<td>
<?php
$sql_sala="SELECT * FROM sala";
$rez_sala= mysqli_query($link,$sql_sala);
echo "<select name=\"sala\" >";
while($sala=mysqli_fetch_array($rez_sala))
{
echo "
<option value=\"".$sala['id_s'];
if ($sala[
'id_s']==$salaa)
echo "\" selected=\"selected\">".$sala['salaa']."</option>\n";
else
echo "\">".$sala['salaa']."</option>\n";
}
echo "</select>";
?><br>
</td>
</tr>
<tr>
<td>Tip</td>
<td>
<?php
$sql_type="SELECT * FROM examen";
$rez_type= mysqli_query($link,$sql_type);
echo "<select name=\"type\" >";
while($type=mysqli_fetch_array($rez_type))
{
echo "
<option value=\"".$type['id_tip'];
if ($type[
'id_tip']==$tip)
echo "\" selected=\"selected\">".$type['tip']."</option>\n";
else
echo "\">".$type['tip']."</option>\n";
}
echo "</select>";
?><br>
</td>
</tr>
<tr>
<td><input name="submit" type="submit" value="Update"/></td>
<td><input name="reset" type="reset" value="Reset"/></td>
</tr>
</table>
</form>
</div>
这是更新:
<?php
include ('conect.php');
$idd=$_POST['idu'];
$a1=$_POST['data'];
$b1=$_POST['year'];
$c1=$_POST['mat'];
$d1=$_POST['proff'];
$e1=$_POST['profff'];
$f1=$_POST['sala'];
$g1=$_POST['type'];
$str = "UPDATE examen_programat
SET data='$a1',
id_ann='$b1',
id_mate='$c1',
id_prof1='$d1',
id_prof2='$e1',
id_sala='$f1',
id_tipp='$g1'
WHERE id='$idd'";
$result=mysqli_query($link,$str);
// if successfully updated.
if($result){
header("Location:tabelex.php");
}
else {
echo "ERROR";
}
?>
我的 examen_programat 表看起来像这样:
id id_ann id_mate id_prof1 id_prof2 id_sala id_tipp
最佳答案
短的开放标签可能会导致这个问题，在你的最后两个td元素。添加php在你的 ? 之后
<td>
<a href="editexamene.php?id=<?php echo $rows['id'];?>"> Edit </a>
</td>
<td>
<a href="deleteexamene.php?id=<?php echo $rows['id']; ?>"> Stergere </a>
</td>
来自 PHP 文档: http://php.net/manual/en/language.basic-syntax.phptags.php
PHP also allows for short open tag
<?(which is discouraged since it is only available if enabled using the short_open_tag php.ini configuration file directive, or if PHP was configured with the --enable-short-tags option).
关于php - 从表结果中获取 id，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/30414098/