数组仅替换值特定键
我有两个数组，在 json 中是这样的（标签）：
[
{
"dates": "2019-07-01",
"c_job": 0
},
{
"dates": "2019-07-02",
"c_job": 0
},
{
"dates": "2019-07-03",
"c_job": 0
},
{
"dates": "2019-07-04",
"c_job": 0
}
]
和来自数据库的其他数据是这样的（数据）：
{
"EXPORT": [
{
"dates": "2019-07-01",
"c_job": 12
},
{
"dates": "2019-07-02",
"c_job": 8
},
{
"dates": "2019-07-04",
"c_job": 11
}
],
"IMPORT": [
{
"dates": "2019-07-03",
"c_job": 11
}
]
}
之后，我尝试使用以下代码替换数组以使用标签使数据精确：
foreach($data as $key => $value) {
$datafinal[$key] = array_replace($label,$value);
}
并且输出是 (datafinal) ：
{
"EXPORT": [
{
"dates": "2019-07-01",
"c_job": 12
},
{
"dates": "2019-07-02",
"c_job": 8
},
{
"dates": "2019-07-04",
"c_job": 11
},
{
"dates": "2019-07-04",
"c_job": 0
}
],
"IMPORT": [
{
"dates": "2019-07-03",
"c_job": 11
},
{
"dates": "2019-07-02",
"c_job": 0
},
{
"dates": "2019-07-03",
"c_job": 0
},
{
"dates": "2019-07-04",
"c_job": 0
}
]
}
我想要的是这样的：
{
"EXPORT": [
{
"dates": "2019-07-01",
"c_job": 12
},
{
"dates": "2019-07-02",
"c_job": 8
},
{
"dates": "2019-07-03",
"c_job": 0
},
{
"dates": "2019-07-04",
"c_job": 11
}
],
"IMPORT": [
{
"dates": "2019-07-01",
"c_job": 0
},
{
"dates": "2019-07-02",
"c_job": 0
},
{
"dates": "2019-07-03",
"c_job": 11
},
{
"dates": "2019-07-04",
"c_job": 0
}
]
}
我尝试过减少和替换数组
aluckdog2回答
-
翻翻过去那场雪
基本上你需要做的是对中的每个顶级数组$data，遍历 0 值数组（我称之为$blank）并查看当前$data数组中是否存在日期。如果是，请复制该值，否则使用空白值：$datafinal = array();foreach ($data as $key => $value) { foreach ($blank as $bkey => $bvalue) { if (($dkey = array_search($bvalue['dates'], array_column($value, 'dates'))) !== false) { $datafinal[$key][$bkey] = $value[$dkey]; } else { $datafinal[$key][$bkey] = $bvalue; } }}echo json_encode($datafinal, JSON_PRETTY_PRINT);输出：{ "EXPORT": [ { "dates": "2019-07-01", "c_job": 12 }, { "dates": "2019-07-02", "c_job": 8 }, { "dates": "2019-07-03", "c_job": 0 }, { "dates": "2019-07-04", "c_job": 11 } ], "IMPORT": [ { "dates": "2019-07-01", "c_job": 0 }, { "dates": "2019-07-02", "c_job": 0 }, { "dates": "2019-07-03", "c_job": 11 }, { "dates": "2019-07-04", "c_job": 0 } ]}3v4l.org 上的演示 -
噜噜哒
我已经使用了旧的和简单的嵌套foreach循环，尽可能简单快捷（代码中的注释）......$template = json_decode('[ { "dates": "2019-07-01", "c_job": 0 }, { "dates": "2019-07-02", "c_job": 0 }, { "dates": "2019-07-03", "c_job": 0 }, { "dates": "2019-07-04", "c_job": 0 } ]', true);$update = json_decode('{ "EXPORT": [ { "dates": "2019-07-01", "c_job": 12 }, { "dates": "2019-07-02", "c_job": 8 }, { "dates": "2019-07-04", "c_job": 11 } ], "IMPORT": [ { "dates": "2019-07-03", "c_job": 11 } ]}', true);和主要代码...// Create blank output array with all entries$output = ["EXPORT" => $template, "IMPORT" => $template];// Loop over update data (from database) - you may need to tweak this for your use caseforeach ( $update as $type => $updateItem ) { // Loop over each set of update values (a row of dates and c_job) foreach ( $updateItem as $updateItem ) { // Locate in empty output array foreach ( $output[$type] as &$item ) { // Same date - update if ( $updateItem['dates'] == $item['dates']) { $item['c_job'] = $updateItem['c_job']; // Stop looking as already updated break; } } }}较短，但涉及更多array_功能，这会创建以日期为键的模板数组，因此您可以使用输入数据中的日期直接更新数据...// Create template array using dates as the array key$template = array_column($template, null, 'dates');$output = ["EXPORT" => $template, "IMPORT" => $template];foreach ( $update as $type => $export ) { foreach ( $export as $export ) { // Use the combination of type (EXPORT) and dates from the update data // to directly update the output (assumes empty value created) $output[$type][$export['dates']]['c_job'] = $export['c_job']; }}// Re-index data to remove dates as keys$output = ["EXPORT" => array_values($output['EXPORT']), "IMPORT" => array_values($output['IMPORT'])];
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