多选框仅通过第一个选定字段更新mysql表中的输出 [英] Multiselect box only updates output in mysql table by the first selected field

问题描述

你好所有:)



我有一个表格应该将数据添加到两个mysql表食谱和recipe_ingredients。后者是一个链接实体。其中一个字段是多选下拉框。当我使用CTRL选择多个托管时，结果只会在下拉框中返回一个选定的选项...我已经在这里工作了几个小时..没有任何建议......我的代码是......



HTML表格

hello all:)

I have a form that supposed to add data to two mysql tables recipes and recipe_ingredients. The latter which is a link entity. One of the fields in the form a multiselect drop down box. When i use CTRL to select multiple entrys the results only return one chosen options in the drop down box...i have been at this for hours..any sugguestions..here is my code

HTML FORM

<table class="form_table">
<tr><form method="POST" onSubmit action="thankyou.php">
</tr><tr>
<td class="ctable_row">Recipe Name:</td>
<td class="ctable_row"><input type="text" name="recipename"></td></tr>
<tr>
<td class="ctable_row">Recipe Author:</td>
<td class="ctable_row"><input type="text" name="recipeauthor"></td>
</tr>
<tr>
<td class="ctable_row">Cook Time (hh:mm:ss): </td>
<td class="ctable_row"><input type="text" name="cooktime"></td>
</tr>
<tr>
<td class="ctable_row">Select From A List of Ingredients<br/>(**hold CTRL to select<br/> multiple values**)</td>
<td class="ctable_row"><select multiple="multiple" name="Ingredients[]">
    <option value="Broccoli">Broccoli</option>
    <option value="Carrots">Carrots</option>
    <option value="Cheese">Cheese</option>
    <option value="Macaroni">Macaroni</option>
    <option value="Oil">Oil</option>
    <option value="Flour">Flour</option>
    <option value="Eggs">Eggs</option>
    <option value="Eggs">Chicken</option>
</select>
</select></td></tr>
<tr>
<td class="ctable_row">Recipe Stages</td>
<td colspan="2"><textarea rows="20" cols="65" name="stages">
 Input Recipe Stages here...
</textarea>
<input type="submit" value="Submit">
</td></tr>
</form>
 </div>
</td>

PHP代码





session_start（）; $ q = mysql_connect（localhost，root，）;



$ db = mysql_select_db（tasteofhome）;







$ recipename = $ _ POST ['recipename'];



$ recipeauthor = $ _ POST ['recipeauthor'];



$ cooktime = $ _ POST ['cooktime'];





$阶段= $ _ POST ['阶段'];



$ own = $ _ POST ['owningredients' ];













$ name = $ _ SESSION ['login_user'];





$ sql2 =SELECT user_id from用户user_name ='$ name';

$ q1 = mysql_query（$ sql2）;





while（$ row = mysql_fetch_array（$ q1））{









$ s =插入食谱（recipe_name，recipe_author，cook_time，stages_description，user_id）值（'$ recipename'，'$ recipeauthor'，'$ cooktime'， $阶段'，'$ row [0]'）;

;





$ sql5 = mysql_query（$ s）;



}

if（$ sql5 === FALSE）{

die （mysql_error（））; // TODO：更好的错误处理

}

if（$ sql5）

{





}



else {

echo未插入数据;



}



foreach（$ _POST ['Ingredients']作为$ selectedOption）{









$ sql =从ingredients_name ='$ selectedOption'的成分中选择ingredients_id;;



$ q = mysql_query（$ sql）; }



while（$ rows = mysql_fetch_array（$ q））



{





$ sql3 =从食谱中选择recipe_id，其中recipe_name ='$ recipename';



$ s = mysql_query（$ sql3）;











echo $ rows [0];



while（$ table2 = mysql_fetch_array（$ s））



{











$ sql2 =插入recipe_ingredients（ingredients_id，recipe_id）值（'$ rows [0]'，'$ table2 [0]'）;

$ w = mysql_query（$ sql2）;

}}

if（$ w）

{

echo数据成功输入数据库
;



}







if（$ w == true）{

header（'refresh：2; url = myre cipes.php'）;



}



？>



返回

推荐答案

q = mysql_connect（localhost ，'root'，''）;

q=mysql_connect("localhost",'root','');

db = mysql_select_db（tasteofhome）;

db=mysql_select_db("tasteofhome");

recipename =

recipename=

这篇关于多选框仅通过第一个选定字段更新mysql表中的输出的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！