Boj*_*mov -3 php mysql sql create-table
$sql = "CREATE TABLE IF NOT EXISTS questions_$username(".
"question_id INT NOT NULL AUTO_INCREMENT, ".
"question MEDIUMTEXT, ".
"answer CHAR(1), ".
"PRIMARY KEY (question_id))";
$retval = mysql_query($sql, $conn) or die(mysql_error());
$sql = "CREATE TABLE IF NOT EXISTS tests_$username(".
"test_id INT NOT NULL AUTO_INCREMENT, ".
"name VARCHAR(30) NOT NULL, ".
"duration INT NOT NULL, ".
"PRIMARY KEY (test_id))";
$retval = mysql_query($sql, $conn) or die(mysql_error());
$sql = "CREATE TABLE IF NOT EXISTS questions_tests_$username(".
"test_id INT NOT NULL, ".
"question_id INT NOT NULL, ".
"FOREIGN KEY (test_id) REFERENCES tests_$username(test_id), ".
"FOREIGN KEY (question_id) REFERENCES questions_$username(question_id), ".
"PRIMARY KEY (test_id, question_id))".
$retval = mysql_query($sql, $conn) or die(mysql_error());
echo "debug";
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前两个表是成功创建的,但第三个表没有成功.它甚至没有任何错误.最后一行被执行.我的数据库中的表数没有限制.
我想你需要重新思考一般模式.
你应该不为每个用户创建新表!
您应该拥有测试,问题,test_questions,答案和用户等表格.
像这样的东西:
tests:
id, name, duration
questions:
id, question
test_questions:
id, test_id, question_id
users:
id, name
answers:
id, test_questions_id, user_id, answer
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然后,您知道哪个用户可以使用一个查询轻松回答哪个问题.