PHP:我没有选择我在表中插入的所有数据,但有些人会帮助我
我很好地在表中插入了数据,它显示在 phpMyadmin 中,但是当我尝试显示数据库中的数据时,只显示一项。我需要帮助。
下面的代码和屏幕截图 下面的
<?php
$sql = "SELECT * FROM nw";
$result = mysqli_query($conn, $sql);
if($result == TRUE){
$count = mysqli_num_rows($result);
if($count > 0){
while($row = mysqli_fetch_assoc($result)){
$incidentId = $row['id'];
$icTypeName = $row['inctype_name'];
$addedBy = $row['added_by'];
$addedOn = $row['added_on'];
}
?>
<tr class="tableData">
<td><?php echo $incidentId;?></td>
<td><?php echo $icTypeName;?></td>
<td><?php echo $addedBy;?></td>
<td><?php echo $addedOn;?></td>
<td><i class="uil uil-edit icon editIcon"></i></td>
</tr>
<?php
} else {
$_SESSION['noData'] = 'No incidents to show!';
}
} else {
die('Failed to Connect!');
}
?>
图片
我希望获取使用“SELECT * FROM nw”显示的数据库表中的所有数据
I inserted the data in the table very well and it's showing in phpMyadmin but when i try to display data from the database only one single item is displayed. I need help.
Code and screenshoots below
<?php
$sql = "SELECT * FROM nw";
$result = mysqli_query($conn, $sql);
if($result == TRUE){
$count = mysqli_num_rows($result);
if($count > 0){
while($row = mysqli_fetch_assoc($result)){
$incidentId = $row['id'];
$icTypeName = $row['inctype_name'];
$addedBy = $row['added_by'];
$addedOn = $row['added_on'];
}
?>
<tr class="tableData">
<td><?php echo $incidentId;?></td>
<td><?php echo $icTypeName;?></td>
<td><?php echo $addedBy;?></td>
<td><?php echo $addedOn;?></td>
<td><i class="uil uil-edit icon editIcon"></i></td>
</tr>
<?php
} else {
$_SESSION['noData'] = 'No incidents to show!';
}
} else {
die('Failed to Connect!');
}
?>
Images below
I expect to get all the data in the database tables displayed with the "SELECT * FROM nw"
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只需将结束的while循环结束的
}
移动到使用变量的代码之后。您当前在输出任何内容之前都要消耗所有结果集,因此您只会看到最后的行数据。Simply move the
}
that ends the while loop to after the code that is using the variables. You are currently consuming all the resultset before outputting anything, so you will only see the last rows data.