PHP中SQL数据表中未显示的所有值
我试图显示从MySQL到PHP的所有数据,如下所示,但只显示1个数据PHP中SQL数据表中未显示的所有值,php,mysql,sql,Php,Mysql,Sql,我试图显示从MySQL到PHP的所有数据,如下所示,但只显示1个数据 <div class="content"> <div class="animated fadeIn"> <div class="row"> <div class="col-md-12"> <div class="card"> <div class="card-header"> <strong class="card-title">Data Ta
<div class="content">
<div class="animated fadeIn">
<div class="row">
<div class="col-md-12">
<div class="card">
<div class="card-header">
<strong class="card-title">Data Table</strong>
</div>
<div class="card-body">
<?php
// Attempt select query execution
$sql = "SELECT * FROM registers";
if($result = mysqli_query($link, $sql)){
if(mysqli_num_rows($result) > 0){
while($row = mysqli_fetch_array($result)){
$test= $row['id'];
}
// Free result set
mysqli_free_result($result);
} else{
echo "No records matching your query were found.";
}
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
// Close connection
?>
<table id="bootstrap-data-table" class="table table-striped table-bordered">
<thead>
<tr>
<th>Name</th>
<th>Position</th>
<th>Office</th>
<th>Salary</th>
</tr>
</thead>
<tbody>
<tr>
<td><?php echo $test; ?></td>
<td><?php echo $name?></td>
<td><?php echo $email?></td>
<td><?php echo $company?></td>
</tr>
</tbody>
</table>
</div>
</div>
</div>
数据表
名称
位置
办公室
薪水
现在的问题是,表中只显示一个数据,如何使用相同的代码显示所有数据?有人能告诉我吗?您的代码有几个问题,您似乎从未设置$name、$email和$company变量,并且您已经循环了一个数组,并在循环后设置了HTML,因此您将只显示一个结果 我马上就来
<table id="bootstrap-data-table" class="table table-striped table-bordered">
<thead>
<tr>
<th>Name</th>
<th>Position</th>
<th>Office</th>
<th>Salary</th>
</tr>
</thead>
<?php
// Attempt select query execution
$sql = "SELECT * FROM registers";
if($result = mysqli_query($link, $sql)) {
if(mysqli_num_rows($result) > 0){
while($row = mysqli_fetch_array($result)) {
$test= $row['id'];
<tr>
<td><?=$row['id']; ?></td>
<td><?=$row['name']; ?></td>
<td><?=$row['email']; ?></td>
<td><?=$row['company']; ?></td>
</tr>
}
mysqli_free_result($result);
} else {
echo "No records matching your query were found.";
}
} else {
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
?>
</table>
名称
位置
办公室
薪水
}
mysqli_免费_结果($result);
}否则{
echo“未找到与您的查询匹配的记录。”;
}
}否则{
echo“错误:无法执行$sql。”.mysqli_错误($link);
}
?>
您多次设置$test
,但是你只能回显一次。@KIKOSoftware但是我怎么能在不回显每个值的情况下多次显示呢?你可以回显while
循环中的表行,从数据库中检索数据。我看不到任何JavaScript。我能建议你去网上阅读几篇教程,开始新的学习吗处理查询结果的概念。如果您只将一件事放入数组中,那么可以从一个小的点开始,它会更快地执行$test[]=$row
而不是array\u push()
,因为您没有调用函数的开销
<?php
// Attempt select query execution
$sql = "SELECT * FROM registers";
$test=array();
if($result = mysqli_query($link, $sql)){
if(mysqli_num_rows($result) > 0){
while($row = mysqli_fetch_array($result)){
array_push($test,$row);
}
// Free result set
mysqli_free_result($result);
} else{
echo "No records matching your query were found.";
}
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
<?php
foreach($test as $record){?>
<tr>
<td><?php echo $test['id']; ?></td>
<td><?php echo $test['name'];?></td>
<td><?php echo $test['email'];?></td>
<td><?php echo $test['company']?></td>
</tr>
<?php } ?>