我正在构建一个php网站，我希望用户创建他的公司。
脚本正在检查用户是否已经注册了任何公司，然后它应该显示是否存在。
如果他没有注册公司，他应该看一张他可以注册的表格。
如果他选择注册公司，脚本将检查任何具有相同名称的公司或插入该行。
我唯一的问题是，当已经有一个具有该名称的公司时，回声不显示。
我已经在代码中写了问题所在。
<?php
$con=mysqli_connect("mysql","USER","PASS","DB");
if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); }
$result_get_companies = mysqli_query($con,"SELECT * FROM companies WHERE userid='". $login_session ."' ORDER BY companyid ASC") or die(mysqli_error());
if (mysqli_num_rows($result_get_companies) >= 1) {
while($row_companies = mysqli_fetch_array( $result_get_companies )) {
$result_get_company_owner = mysqli_query($con,"SELECT username FROM users WHERE userid='". $login_session ."'") or die(mysqli_error());
$company_owner = mysqli_fetch_assoc($result_get_company_owner);
echo 'THIS WORKS';
}
} else {
if (isset($_POST['create_first_company']) && !empty($_POST['company_name'])) {
$company_name_unsafe = mysqli_real_escape_string($con, $_POST['company_name']);
$company_name = preg_replace("/[^a-zA-Z0-9\s]/","",$company_name_unsafe );
$check_companies = "SELECT companyid FROM companies WHERE company_name='". $company_name ."'";
$what_to_do_companies = mysqli_query($con,$check_companies);
if (mysqli_num_rows($what_to_do_companies) != 0) {
echo 'THIS DOESNT WORK
It does register that is should go here
because it does not insert new row.
and when the value is = 0 it does go
to else ELSE below and insert row.';
} else {
$result_create_company = mysqli_query($con,"INSERT INTO companies (companyname)
VALUES ('". $login_session ."')")
or die(mysqli_error());
echo 'THIS WORKS';
}
} else {
echo 'THIS WORKS!';
}
}
?>