我试图根据所选员工的价值取得相关的津贴价格。当我选择一名员工时,会生成津贴类型,但每种津贴类型都会回显相同的价格。
我尝试了以下代码,但无法找到错误。你能帮我找一下代码中的错误吗?
<div class="form-group">
<label class="control-label col-lg-4" for="advanceAmount">Allowances: </label>
<div class="col-lg-3">
<?php
for($i = 0; $i < count($b); $i++){
echo '<input type="text" class="form-control input-sm margin-bottom" name="allowance_type" value="' . $b[$i] . '">';
}
?>
</div>
<div class="col-lg-3">
<?php
$i = 0;
$s = mysqli_query($connection,"select * from allowances_payroll where allowance_name = '$b[$i]'");
$r = mysqli_fetch_assoc($s);
for($i = 0; $i < count($b); $i++){
echo '<input type="text" class="form-control input-sm margin-bottom" value="' . $r['allowance_amount'] . '">';
}
?>
</div>
</div>
$ b是来自数据库的爆炸津贴类型。
if($result1){
while($record = mysqli_fetch_array($result1)){
$empId = $record['emp_id'];
$department = $record['dep_name'];
$salary = $record['emp_basicSalary'];
$designation = $record['emp_classification'];
$allowance = $record['allowances_types'];
$b = explode(" ,", $allowance);
}
}
由于
答案 0 :(得分:0)
你在错误的地方循环。 将您的代码更改为:
for($i = 0; $i < count($b); $i++){
$s = mysqli_query($connection,"select * from allowances_payroll where allowance_name = '{$b[$i]}'");
$r = mysqli_fetch_assoc($s);
echo '<input type="text" class="form-control input-sm margin-bottom" value="' . $r['allowance_amount'] . '">';
}
}
还要注意$b[$i]
周围的{},或者你可以像其他两个一样连接。