Question

php code

<?php
$hostname="localhost";
$username="root";
$password="tiger";

  /* @var $dbhandle type */
 $dbhandle = \mysqli_connect($hostname, $username, $password) 
  or die("Unable to connect to MySQL");

    /* @var $select type */
       $select= \mysqli_select_db($dbhandle,"sample")
     or mysqli_error($dbhandle);
$tnumber=(\filter_input(INPUT_POST,'tnumber'));
$capacity=(\filter_input(\INPUT_POST,'capacity'));
$status=(\filter_input(\INPUT_POST,'status'));
$sql1="select * from tablecheck ";
$res=mysqli_query($dbhandle,$sql1);
$row=  mysqli_fetch_array($res); 
if($row['status'] === 'Booked')
{
echo "Please select another table";
}
else{
 $sql="update tablecheck set status='Booked' where tnumber='$tnumber' ";
/* @var $result type */
$result= \mysqli_query($dbhandle,$sql) or die(\mysqli_error($dbhandle));
echo "success";

  mysqli_close($dbhandle); 
}

总是当我在文本框中输入表格编号时，即使状态未被预订，也会执行，因为请选择另一个表格，我的其他部分无效。如何克服此错误..

Answer 1

首先检查您是否获得$ row [＆＃39; status＆＃39;]的值为＆＃34; Booked＆＃34;

$sql1="select * from tablecheck ";

在此声明中，您需要检查未预订状态的行，您返回的选择查询可能无法始终返回正确的结果

Answer 2

更改($row['status'] === 'Booked')

的($row['status'] == 'Booked')

else部分不能在php中工作

2 个答案: