今天我从我的数据库中删除了make table,但不知何故,当我运行tis代码时,它会永远加载网站= \ $pismenka[0] to [25] are A, B, C .....
Databse 100%在我调用它时工作但是当我使用数组时它不是= \
Echo仅用于测试是否找到数据以及代码是否到达最终
问题短缺:网站永远不会加载所以没有错误= \只是whitesmoke屏幕
问题部分:
while($db_data = mysqli_fetch_assoc(mysqli_query($connect_to_db, "SELECT * FROM anime WHERE a_name LIKE '$start1%'"))) {
echo "aaaaaaaaaaaaaaa";
}
整个代码:
for ($i = 0; $i < count($pismenka); $i++) {
echo "<div id='pismenko' class='text-center'>$pismenka[$i]</div>";
$start1 = $pismenka[$i];
echo "
<table class=\"table table-striped table-hover\">
<thead>
<tr>
<th style=\"width: 30%;\">Názov anime</th>
<th style=\"width: 10%;\">Rok vydania</th>
<th style=\"width: 10%;\">Preložené</th>
<th style=\"width: 10%;\">Hodnotenie</th>
<th style=\"width: 10%;\">Preklad</th>
<th style=\"width: 10%;\">Stav</th>
</tr>
</thead>
<tbody>
";
while($db_data = mysqli_fetch_assoc(mysqli_query($connect_to_db, "SELECT * FROM anime WHERE a_name LIKE '$start1%'"))) {
echo "aaaaaaaaaaaaaaa";
}
echo "</tbody></table>";
}
答案 0 :(得分:2)
您有一个无限循环,因为您不断重新执行相同的查询并获得相同的结果。
执行查询一次:
$result = mysqli_query($connect_to_db, "SELECT * FROM anime WHERE a_name LIKE '$start1%'");
然后遍历该查询的结果:
while($db_data = mysqli_fetch_assoc($result)) {
echo "aaaaaaaaaaaaaaa";
}