SQL連接查詢問題

Question

我有一個名為tutor_signup_form的mysql表看起來像這樣

id    feerange    gender    name
1     20          male      alex      
2     10          female    Robin
3     15          male      Babu

現在，我有一個html搜索表單，其中沒有幾個下拉選定的字段，並且存在許多Chckbox字段。 我想使用sql INNER JOIN從多個表中獲取搜索結果。 我的INNER JOIN查詢看起來像這樣

// search query
$query =  mysql_query("SELECT tutor_signup_form . *, tutor_signup_edu_psle . *,
tutor_signup_edu_olevel . *, tutor_signup_edu_alevel . *, tutor_signup_edu_diploma . *,
tutor_signup_edu_degree . *, tutor_signup_edu_masters . *, tutor_signup_edu_psd . *,
tutor_signup_level_primary . *, tutor_signup_level_olevel . *, tutor_signup_level_alevel . *,
tutor_signup_level_int . *, tutor_signup_level_uni . *, tutor_signup_level_music . *,
tutor_signup_level_lang . *, tutor_signup_level_com . *, tutor_signup_pre_central . *,
tutor_signup_pre_east . *, tutor_signup_pre_west . *, tutor_signup_pre_south . *,
tutor_signup_pre_north . *, tutor_signup_pre_ne . *, tutor_signup_pre_nw . *  FROM tutor_signup_form

INNER JOIN tutor_signup_edu_psle ON tutor_signup_form.tutor_id = tutor_signup_form.tutor_id     
INNER JOIN tutor_signup_edu_olevel ON tutor_signup_form.tutor_id = tutor_signup_edu_olevel.tutor_id
INNER JOIN tutor_signup_edu_alevel ON tutor_signup_form.tutor_id = tutor_signup_edu_alevel.tutor_id 
INNER JOIN tutor_signup_edu_diploma ON tutor_signup_form.tutor_id = tutor_signup_edu_diploma.tutor_id
INNER JOIN tutor_signup_edu_degree ON tutor_signup_form.tutor_id = tutor_signup_edu_degree.tutor_id
INNER JOIN tutor_signup_edu_masters ON tutor_signup_form.tutor_id = tutor_signup_edu_masters.tutor_id
INNER JOIN tutor_signup_edu_psd ON tutor_signup_form.tutor_id = tutor_signup_edu_psd.tutor_id
INNER JOIN tutor_signup_level_primary ON tutor_signup_form.tutor_id = tutor_signup_level_primary.tutor_id
INNER JOIN tutor_signup_level_olevel ON tutor_signup_form.tutor_id = tutor_signup_level_olevel.tutor_id
INNER JOIN tutor_signup_level_alevel ON tutor_signup_form.tutor_id = tutor_signup_level_alevel.tutor_id
INNER JOIN tutor_signup_level_int ON tutor_signup_form.tutor_id = tutor_signup_level_int.tutor_id
INNER JOIN tutor_signup_level_uni ON tutor_signup_form.tutor_id =  tutor_signup_level_uni.tutor_id
INNER JOIN tutor_signup_level_music ON tutor_signup_form.tutor_id = tutor_signup_level_music.tutor_id
INNER JOIN tutor_signup_level_lang ON tutor_signup_form.tutor_id = tutor_signup_level_lang.tutor_id
INNER JOIN tutor_signup_level_com ON tutor_signup_form.tutor_id =  tutor_signup_level_com.tutor_id
INNER JOIN tutor_signup_pre_central ON tutor_signup_form.tutor_id = tutor_signup_pre_central.tutor_id
INNER JOIN tutor_signup_pre_east ON tutor_signup_form.tutor_id = tutor_signup_pre_east.tutor_id
INNER JOIN tutor_signup_pre_west ON tutor_signup_form.tutor_id = tutor_signup_pre_west.tutor_id
INNER JOIN tutor_signup_pre_south ON tutor_signup_form.tutor_id = tutor_signup_pre_south.tutor_id
INNER JOIN tutor_signup_pre_north ON tutor_signup_form.tutor_id =  tutor_signup_pre_north.tutor_id
INNER JOIN tutor_signup_pre_ne ON tutor_signup_form.tutor_id = tutor_signup_pre_ne.tutor_id
INNER JOIN tutor_signup_pre_nw ON tutor_signup_form.tutor_id = tutor_signup_pre_nw.tutor_id 

WHERE tutor_signup_form.feerange <= '$budget'");  

例如，我選擇的預算字段=20。因此，我的查詢應該是tutor_signup_form僅乘積3個結果，但它會產生24個結果。 我不明白為什么會產生24個結果。 你能修好它嗎 ？

更新：

Answer 1

正如戴德曼所說，笛卡爾式...

您的第一個內部聯接是表名稱的副本/粘貼

INNER JOIN tutor_signup_edu_psle 
   ON tutor_signup_form.tutor_id = tutor_signup_form.tutor_id     

本來應該

INNER JOIN tutor_signup_edu_psle 
   ON tutor_signup_form.tutor_id = tutor_signup_edu_psle.tutor_id     

我沒有讀過去，但與此同時

此外，您所有的聯接都為“內部聯接”。 如果某人在其中一個表中沒有記錄（例如音樂），但是在計算機中進行了輔導，並且有人在搜索計算機，則不會返回這些記錄。 我建議所有人員都改為LEFT JOIN，因此，如果該記錄存​​在於補習子公司級別，那么您會很好，並且不會失去它。

Answer 2

如果我做對了，則需要檢查您的where子句。我認為聯接所涉及的表的笛卡爾積。 笛卡爾乘積結果集的大小是第一個表中的行數乘以第二個表中的行數