这一个工作正常#1,但是当我尝试在第二个表上关联其他表时,它给出了一个错误
注意:未定义的索引:第30行的C:\ xampp \ htdocs \ hyukies \ public \ samples.php中的recipe.id
注意:第31行的C:\ xampp \ htdocs \ hyukies \ public \ samples.php中的未定义索引:recipe.ingredientid RecipeID:IngredientID:
注意:未定义的索引:第30行的C:\ xampp \ htdocs \ hyukies \ public \ samples.php中的recipe.id
注意:第31行的C:\ xampp \ htdocs \ hyukies \ public \ samples.php中的未定义索引:recipe.ingredientid RecipeID:IngredientID:
注意:未定义的索引:第30行的C:\ xampp \ htdocs \ hyukies \ public \ samples.php中的recipe.id
注意:第31行的C:\ xampp \ htdocs \ hyukies \ public \ samples.php中的未定义索引:recipe.ingredientid RecipeID:IngredientID:
<?php
$sql = mysqli_query($connection, "SELECT id, location FROM location");
$userinfo = array();
while ($row_user = mysqli_fetch_assoc($sql))
$userinfo[] = $row_user;
foreach ($userinfo as $user) {
echo "ID: {$user['id']}<br />"
. "Location: {$user['location']}<br /><br />";
}
?>
这是一个无法正常工作的错误问题,请帮忙......
<?php
$sql_2 = mysqli_query($connection, "SELECT recipe.id, recipe.ingredientid, ingredients.id, ingredients.quantity, ingredients.`name` FROM recipe INNER JOIN ingredients ON ingredients.id = recipe.ingredientid");
$userinfo_2 = array();
while ($row_user_2 = mysqli_fetch_assoc($sql_2))
$userinfo_2[] = $row_user_2;
foreach ($userinfo_2 as $user_2) {
echo "RecipeID: {$user_2['recipe.id']}<br />"
. "IngredientID: {$user_2['recipe.ingredientid']}<br /><br />";
}
?>
删除查询中“名称”的反勾号
$sql_2 = mysqli_query($connection, "SELECT recipe.id, recipe.ingredientid, ingredients.id, ingredients.quantity, ingredients.`name` FROM recipe INNER JOIN ingredients ON ingredients.id = recipe.ingredientid");
应该
$sql_2 = mysqli_query($connection, "SELECT recipe.id, recipe.ingredientid, ingredients.id, ingredients.quantity, ingredients.name FROM recipe INNER JOIN ingredients ON ingredients.id = recipe.ingredientid");
另外,从中删除“食谱” $user_2['recipe.id']
。应该是$user2['id']
,但不清楚您要获得哪个ID。要独立选择它们,可以在查询中使用“ AS”为它们提供唯一的标识符。
SELECT recipe.id AS id1, ingedients.id AS id2
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我来说两句