Question

我需要从网站root@amnesia:/home/amnesia# ssh root@11.11.11.11 FATAL: Failed to determine SOCKS server. ssh_exchange_identification: Connection closed by remote host root@amnesia:/home/amnesia# ssh -v root@11.11.11.11 OpenSSH_6.7p1 Debian-5+deb8u1, OpenSSL 1.0.1k 8 Jan 2015 debug1: Reading configuration data /etc/ssh/ssh_config debug1: /etc/ssh/ssh_config line 4: Applying options for * debug1: Executing proxy command: exec /usr/local/lib/connect-socks 11.11.11.11 22 debug1: permanently_set_uid: 0/0 debug1: identity file /root/.ssh/id_rsa type 1 debug1: key_load_public: No such file or directory debug1: identity file /root/.ssh/id_rsa-cert type -1 debug1: key_load_public: No such file or directory debug1: identity file /root/.ssh/id_dsa type -1 debug1: key_load_public: No such file or directory debug1: identity file /root/.ssh/id_dsa-cert type -1 debug1: key_load_public: No such file or directory debug1: identity file /root/.ssh/id_ecdsa type -1 debug1: key_load_public: No such file or directory debug1: identity file /root/.ssh/id_ecdsa-cert type -1 debug1: permanently_drop_suid: 0 debug1: key_load_public: No such file or directory debug1: identity file /root/.ssh/id_ed25519 type -1 debug1: key_load_public: No such file or directory debug1: identity file /root/.ssh/id_ed25519-cert type -1 debug1: Enabling compatibility mode for protocol 2.0 debug1: Local version string SSH-2.0-OpenSSH_6.7p1 Debian-5+deb8u1 FATAL: Failed to determine SOCKS server. ssh_exchange_identification: Connection closed by remote host获取信息，但只有用户登录才能访问此页面。

因此，我使用以下脚本登录用户，然后从页面获取信息，但是我想知道脚本是否格式正确。

@ManyToMany(cascade = CascadeType.ALL)
@JoinTable(name = "INGREDIENT_RECIPE_REL",
        joinColumns = {
            @JoinColumn(name = "irr_rcp_id", referencedColumnName = "rcp_id")
        },
        inverseJoinColumns = {
            @JoinColumn(name = "irr_ing_ingredient", referencedColumnName = "ing_ingredient"),
            @JoinColumn(name = "irr_ing_acc_identifier", referencedColumnName = "ing_acc_identifier")
        })

Answer 1

首先，不要混用jQuery和JavaScript。如果你要做一个特定的任务，那就是这样或那样。与代码的语法保持一致。其次，代码很好，但是你必须通过将open()的第三个参数设置为false来使代码同步。如果您想保持相同的jQuery语法，请将open()转换为$.post()并登录。然后，在回调中，您可以向您的API发出请求。

xhttp将用户名和密码发送到url

1 个答案: