我是mysqli的新手,并且mysqli的结果有问题。不幸的是,我只得到一个结果。当我把查询放入phpMyAdmin时,它会出现三个结果。我相信相关的代码是在这里和我打电话是一个错误:在mysqli中循环显示结果
$connection = new mysqli($host, $databaseUsername, $databasePassword, $database);
if ($connection->connect_errno > 0) {
die ('Unable to connect to database [' . $connection->connect_error . ']');
}
$sql = "SELECT clientId, studentFirstName, studentLastName
FROM clients
WHERE (studentEmail = '$postEmail') OR (parentEmail = '$postEmail');";
if (!$result = $connection->query($sql)) {
die ('There was an error running query[' . $connection->error . ']');
}
echo '<select class = "toolbarDropdown" id = "toolbarDropdown-MultipleAccounts">';
while ($row = $result->fetch_array()) {
echo '<option value="'.$row["clientId"].'>'.$row["studentFirstName"].' '.$row["studentLastName"].'</option>';
}
echo '</select>';
您可以添加您的实际查询吗? – DACrosby
我们需要查看查询。 – busypeoples
@ DouglasA.Crosby好的,我添加了SQL查询。后来在代码中,我回显$ sql,然后复制并粘贴到phpMyAdmin中。在phpMyAdmin中,这导致了三个结果。 – radleybobins