我几乎完成了我的程序,这是一个简单的状态发布系统。如果用户输入的状态不在数据库内部,则尝试回显错误消息很困难。
我见过MySQLi If condition alert msg not showing the output和ECHO not showing in HTML from PHP,但它们都不能解决我的问题。
我已经尝试将echo语句在超链接之后移至,以查看是否可以解决该问题,但事实并非如此。我也尝试过在工作的else块中添加echo语句。
//Query's used to check if their is data in the 'status' table and if the table 'status' exists
$query = "select * from status where status like '$status%'";
$check_query = "DESC status";
$result = mysqli_query($conn, $check_query);
// checks if the execution was successful
if(!$result) {
//Links to return home if the query's fail.
echo '<br>';
echo '<p>Status does not exist in the database</p>';
echo '<a href="http://mwx9073.cmslamp14.aut.ac.nz/assign1/index.html">Click here to return home</a>';
echo '<br>';
echo '<a href="http://mwx9073.cmslamp14.aut.ac.nz/assign1/searchstatuform.html">Click here to search again</a>';
} else {
//Displaying the status information if the operation is a success
$result = mysqli_query($conn, $query);
while ($row = mysqli_fetch_assoc($result)){
echo "<br> Status: ".$row["status"].
"<br> Status Code: ".$row["code"].
"<br> Share: ".$row["share"].
"<br> Date Posted: ".$row["post_date"].
"<br> Permission: ".$row["permission"];
echo '<br>';
}
echo '<br>';
echo '<br>';
echo '<a href="http://mwx9073.cmslamp14.aut.ac.nz/assign1/index.html">Click here to return home</a>';
echo '<br>';
echo '<a href="http://mwx9073.cmslamp14.aut.ac.nz/assign1/searchstatusform.html">Click here to search again</a>';
// Frees up the memory, after using the result pointer
mysqli_free_result($result);
}
mysqli_close($conn);
我希望echo "<p>Status does not exist in the database</p>";
可以打印,但是仅显示超链接,并且前面没有echo语句。