Kri*_*lev 3 invoice laravel phpword phpoffice
我有 phpword 表的问题。

我有下表,我想克隆第一个表行并替换其中的信息。到目前为止,我没有任何进展。我使用getVariables()方法从文档中获取所有变量并循环遍历它们。我已经检查过该值是否是数组,如果是它属于该行。我按照以下方式构建了数据
Collection {#971 ?
#items: array:12 [?
"ticket_id" => array:1 [?
0 => 7
0 => 6
]
"ticket_number" => array:2 [?
0 => "157-12313121321"
1 => null
]
"price_offered_bgn" => array:2 [?
0 => 978.0
1 => 196.0
]
"ticket_is" => array:1 [?
0 => "Requested"
]
"departure_date" => array:2 [?
0 => "2020-10-20 00:00:00"
1 => "2020-01-29 00:00:00"
]
"return_date" => array:2 [?
0 => "2020-10-29 00:00:00"
1 => null
]
"company_address" => array:1 [?
0 => "ADDRESS"
]
"company_bulstat" => array:1 [?
0 => ""
]
"company_dds_number" => array:1 [?
0 => "BG 104023232353"
]
"mol" => array:1 [?
0 => "Gleichner"
]
"first_name" => array:2 [?
0 => "Araceli"
1 => "Francisca"
]
"last_name" => array:2 [?
0 => "Gleichner"
1 => "Schmitt"
]
]
}
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在我尝试克隆变量并插入值后,我得出了以下结果
array:4 [?
0 => "TICKET_NUMBER"
1 => "FIRST_NAME"
2 => "LAST_NAME"
3 => "DEPARTURE_DATE"
]
array:9 [?
0 => "FIRST_NAME#1"
1 => "LAST_NAME#1"
2 => "DEPARTURE_DATE#1"
3 => "RETURN_DATE#1"
4 => "TICKET_NUMBER#2"
5 => "FIRST_NAME#2"
6 => "LAST_NAME#2"
7 => "DEPARTURE_DATE#2"
8 => "RETURN_DATE#2"
]
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而这个错误
Can not clone row, template variable not found or variable contains markup.
at TemplateProcessor->cloneRow('${FIRST_NAME}', 2)
如果您对我如何克隆这一行并在其中插入值有任何想法,我将不胜感激。
问题解决了。我已经制作了这样的表格结构
+-----------+----------------+
| ${row} | ${Item} |
| | +
| | ${ItemInfo} |
+-----------+----------------+
+-----------+----------------+
| ${row#1} | ${Item} |
| | +
| | ${ItemInfo} |
+-----------+----------------+
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我正在使用PhpWord DocscloneRow('ROW', 2)方法
克隆该行,这为我提供了 2 个 ROW 副本,我可以使用这些副本并添加其中的每个副本。这样我循环遍历它们并用这样的实际值替换占位符#INDEX
foreach ($fields as $key => $value) {
$this->wordFile->setValue(strtoupper($key) . '#' . $index, $value);
$this->wordFile->setValue('ROW#' . $index, $index);
}
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该KEY变量是字段名,然后我将串联#INDEX到它。
克隆的行以索引 1 开始(#1、#2、#3 等等...)。
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