我遇到了一个奇怪的问题。我正在为MySQL数据库开发一个web界面，并尝试使用PHP从中获取信息。简而言之，当我不在其他数据库中时，我可以很好地从一些数据库中检索信息。
$userList = mysql_query("SELECT * FROM myTable");
while ($userInfo = mysql_fetch_array($userList) )
{
echo "<p>Name = " . $userInfo["name"] . ". Password = " . $userInfo["password"] . ".</p>";
}
这部分只是一个测试，它运行良好。我得到了数据库中每个人的名字和密码。但当我尝试用这种方式做时，我会遇到错误。考虑
while ( ($userInfo = mysql_fetch_array($userList) ) && (!$found) )
{
echo "The current username is " . $userInfo["name"] . ". <ul>";
if ($username == $userInfo["name"])
{
$found = true;
if ($password == $userInfo['password'])
{
echo "The password you entered, " . $userInfo['password'] . " was correct!";
//things which we do once the account is confirmed
}
else //the username is right but the password is wrong.
{
"The password didn't match, though!";
}
}
else //this username isn't the right one.
{
echo "<p>$username does not match " . $userInfo['name'] . ".";
}
echo "</ul>";
}
具体地说，$userInfo["name"]和$userInfo["password"]字符在第二块代码中完全不返回任何内容，而在第一块代码中，它们似乎工作得很好。我不明白为什么两者之间有区别。
如果我能得到任何帮助或建议，我们将不胜感激。
编辑：对于那些想要完整代码的人，它就在这里。
<head>
<title>My Web Interface</title>
</head>
<?php
if (!($_POST["go"]))
{
?><h1>Hello World!</h1>
<form action="test.php" method="post">
Username: <input type="text" name="username" /> <br />
Password: <input type="text" name="password" /> <br />
<input type="submit" name="go" />
</form>
<?php
}
else //the user has submitted: in which case, we check if the details match any user info in the database
{
$username = $_POST["username"];
$password = $_POST["password"];
//the database info variables
$hostname = "myhostname";
$dbUsername = "myusername";
$dbPassword = "mypassword";
echo "<p>You entered the username and password combo of '$username' and '$password'.</p>";
$connect = mysql_connect($hostname, $dbUsername, $dbPassword) or die ("Unable to connect to MySQL");
//test for the connection's presence. Every time so far it's returned True.
if ($connect)
{
echo "Got it!";
}
else
{
echo "Don't got it!";
}
//echo "<p>My username is " . $dbUsername ", my hostname is " . $hostname . " and my password is " . $dbPassword . ".</p>";
$selected = mysql_select_db("myDatabase",$connect)
or die("Could not select examples");
$userList = mysql_query("SELECT * FROM testUsers;");
/****
* This part tests to show a connection between the user and the database.
* It should return a list of users and the rights they have.
***
*/
$found = false; //how we terminate the loop.
//echo "<ul>";
while ($userInfo = mysql_fetch_array($userList) )
{
echo "<p>Name = " . $userInfo["name"] . ". Password = " . $userInfo["password"] . ".</p>";
if ($userInfo["password"] == $password)
{
echo "<p>The passwords match and are both $password!</p>";
}
else
{
echo "<p>$password does not match with " . $userInfo["password"] . "!</p>";
}
}
while ( ($userInfo = mysql_fetch_array($userList) ) && (!($found)) )
{
echo "The current username is " . $userInfo["name"] . ". <ul>";
if ($username == $userInfo["name"])
{
$found = true;
echo "<p>We found you in the database, " . $userInfo['name'] . ". Now to test your password.</p>";
if ($password == $userInfo['password'])
{
echo "<p>The password you entered, " . $userInfo['password'] . " was correct!</p>";
//now show the table's contents
$register = mysql_query("SELECT * FROM myTable;");
while ($col = mysql_fetch_array($register) )
{
echo "<li>Tag: " . $col['service_tag'] . "</li>";
}
}
else //the username is right but the password is wrong.
{
echo "The password didn't match, though!";
}
}
else //this username isn't the right one.
{
echo "<p>$username does not match " . $userInfo['name'] . ".";
}
echo "</ul>";
}
/*
*Test code: trying to output the testUsers info without the conditions.
*/
if (!$found)
{
echo "<p>We could not find you in the database. Did you enter your username correctly?</p>";
}
echo "</ul>";
mysql_close($connect);
}
?>
第二版：有些人指出，此演示文稿的密码非常不安全，我同意这一点——这根本不是网站的最终代码。我只是想在遇到这个问题时测试一下连接。
完成第一个块后，光标位于$userList的末尾。因此，在第二块中，没有更多内容可供读取
$userInfo = mysql_fetch_array($userList)
尝试在第二个块之前包含以下语句，将光标移回第一个项目：
mysql_data_seek($userList, 0);
或更好：
if (!mysql_data_seek($userList, 0))
{
echo "You can't go back (seek): " . mysql_error() . "'n";
}
else
{
// Block 2
}
我认为您没有向我们展示完整的代码。您必须在某个地方设置数据库的凭据，这是在您将值设置为$username之前还是之后？也许您也将$username用于数据库连接？
这是因为while循环中的(!($found)。它的最终状态必须是真的，循环才能采取行动，但它一直是假的，因为你要求它不是假的，但它确实是。因此，它不会起作用，因为条件不满足。试着从中删除!并测试它。如果你不能这样做，试着使用这个：
do {
//your code here to motion
}
while()// your conditions
在这种情况下，只要while中的状态为真，do就会继续工作。