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我在第 50 行收到一个错误,提示有意外的 ';' 在我的代码中。
if(mysql_num_rows($result) > 0) { mysqli_query($con, "UPDATE COOL_USERS SET COUNT = COUNT + 1 WHERE USERNAME = '" .mysql_real_escape_string($user). "'"; }
第 50 行是 WHERE USERNAME... 行。谁能发现它?
你)不见了mysqli_query
)
mysqli_query
if(mysql_num_rows($result) > 0) { mysqli_query($con, "UPDATE COOL_USERS SET COUNT = COUNT + 1 WHERE USERNAME = '" .mysql_real_escape_string($user). "'"); }