【发布时间】:2021-08-18 14:28:00
【问题描述】:
我尝试过的是使用 if 语句和 switch 语句,但它仍然无法正常工作,因为它进入了默认值:“出现问题”,我已经编写了代码,但我不确定出了什么问题
<form action="index.php" method="post" id="formSelect">
<div id="chosen">
<label for="pkCountry" class="blue">Choose a country:</label>
<select name="chooseCountry" id="choose">
<option value="start">select...</option>
<option value="Australia">Australia</option>
<option value="India">India</option>
<option value="Pakistan">Pakistan</option>
<option value="England">England</option>
<option value="South Africa">South Africa</option>
<option value="Sri Lanka">Sri Lanka</option>
<option value="Bangladesh">Bangladesh</option>
<option value="West Indies">West Indies</option>
</select>
<?php
$db = new SQLite3('cricket.db');
if (!$db) {
echo $db->lastErrorMsg();
}
else {
switch (isset($_POST['formCountry'])){
case 'India':
$command = $db->query("SELECT * FROM batsman WHERE country = 'India';");
$db->exec($command);
break;
case 'Australia':
$command1 = $db->query("SELECT * FROM batsman WHERE country = 'Australia';");
$db->exec($command1);
break;
case 'England':
$command2 = $db->query("SELECT * FROM batsman WHERE country = 'England';");
$db->exec($command2);
break;
case 'Pakistan':
$command3 = $db->query("SELECT * FROM batsman WHERE country = 'Pakistan';");
$db->exec($command3);
break;
case 'Bangladesh':
$command4 = $db->query("SELECT * FROM batsman WHERE country = 'Bangladesh';");
$db->exec($command4);
break;
case 'Sri Lanka' :
$command5 = $db->query("SELECT * FROM batsman WHERE country = 'Sri Lanka';");
$db->exec($command5);
break;
case 'South Africa':
$command6 = $db->query("SELECT * FROM batsman WHERE country = 'South Africa';");
$db->exec($command6);
break;
case 'West Indies':
$command7 = $db->query("SELECT * FROM batsman WHERE country = 'West Indies';");
$db->exec($command7);
break;
default:
echo "<br>something went wrong</br>";
$db->close();
}
}
?>
【问题讨论】: