我有来自多个表多记录的 json 文件。 json 文件是从本地 mysql 数据库创建的，我还有另一个具有相同模式的在线数据库。但是每个表都没有相同数量的字段和相同的数据/值。但是一些字段名称对于所有表都是相同的(id、added_on、last_updated)。我想通过执行单个 php 文件将这些 json 文件值导入在线数据库。
json文件是这样的:
[
{
"tableName":"table_Name_1",
"rows":[
{
"t1column1":"valuet1row11",
"t1column2":"valuet1row12",
"t1columnx":"valuet1row1x"
},
{
"t1column1":"valuet1row21",
"t1column2":"valuet1row21",
"t1columnx":"valuet1row2x"
},
{
"t1column1":"valuet1rowx1",
"t1column2":"valuet1rowx2",
"t1columnx":"valuet1rowxx"
}
]
},
{
"tableName":"table_Name_2",
"rows":[
{
"t2column1":"valuet2row11",
"t2column2":"valuet2row12",
"t2columnx":"valuet2row1x"
},
{
"t2column1":"valuet2row21",
"t2column2":"valuet2row22",
"t2columnx":"valuet2row2x"
},
{
"t2column1":"valuet2rowx1",
"t2column2":"valuet2rowx2",
"t2columnx":"valuet2rowxx"
}
]
},
{
"tableName":"table_Name_n",
"rows":[
{
"tncolumn1":"valuetnrow11",
"tncolumn2":"valuetnrow12",
"tncolumnx":"valuetnrow1x"
},
{
"tncolumn1":"valuetnrow21",
"tncolumn2":"valuetnrow22",
"tncolumnx":"valuetnrow2x"
},{
"tncolumn1":"valuetnrowx1",
"tncolumn2":"valuetnrowx2",
"tncolumnx":"valuetnrowxx"
}
]
},
]
以下php代码用于将json文件记录导入到数据库中的单个表中。 (这里的single_table.json只包含单表记录)
<?php
try
{
$connect = mysqli_connect("localhost", "fmart", "password", "mart_dbsync");
$query = '';
$table_data = '';
$filename = "single_table.json";
$data = file_get_contents($filename);
$array = json_decode($data, true);
foreach($array as $row)
{
$query .= "INSERT INTO purchases(id, invoicenum, supplier, stock_keeper, counter, added_by, is_deleted, description, is_opening_stock, department, added_on, last_updated) VALUES ('".$row["id"]."', '".$row["invoicenum"]."', '".$row["supplier"]."', '".$row["stock_keeper"]."', '".$row["counter"]."', '".$row["added_by"]."', '".$row["is_deleted"]."', '".$row["description"]."', '".$row["is_opening_stock"]."', '".$row["department"]."', '".$row["added_on"]."', '".$row["last_updated"]."') ON DUPLICATE KEY UPDATE invoicenum='".$row["invoicenum"]."', supplier='".$row["supplier"]."', stock_keeper='".$row["stock_keeper"]."', counter='".$row["counter"]."', added_by='".$row["added_by"]."', is_deleted='".$row["is_deleted"]."', description='".$row["description"]."', is_opening_stock='".$row["is_opening_stock"]."', department='".$row["department"]."', added_on='".$row["added_on"]."', last_updated='".$row["last_updated"]."';";
}
mysqli_multi_query($connect, $query);
echo "<h1>All purchases appended </h1>";
}
catch(Exception $e)
{
echo $e->getMessage();
}
?>
在上面的 php 代码中，表名是硬编码在 INSERT 语句中的。但是通过使用新的json(如上面的json格式)，它包含超过25个表，表名应该从json文件中获取。
这是回显的结果:
contacts
Insert query:
INSERT INTO contacts
(First_Name, Last_Name, Company, Business_Phone, Email_Address)
VALUES
('Dave','Frank','Company1','0115 999999','zvv@zz.com'),
('Dave','Blogs','Company2','0115 888888','zvv@zz.com'),
('David','frank','Company3','0115 777777','zvv@zz.com')
Error: 1
INSERT INTO contacts (First_Name, Last_Name, Company, Business_Phone, Email_Address) VALUES ('Dave','Frank','Company1','0115 999999','zvv@zz.com'), ('Dave','Blogs','Company2','0115 888888','zvv@zz.com'), ('David','frank','Company3','0115 777777','zvv@zz.com')
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1contacts_old
Insert query:
INSERT INTO contacts_old
(First_Name, Last_Name, Company, Business_Phone, Email_Address)
VALUES
('Dave','Frank','Company1','0115 999999','zvv@zz.com'),
('Dave','Blogs','Company2','0115 888888','zvv@zz.com'),
('David','frank','Company3','0115 777777','zvv@zz.com')
Error: 1
INSERT INTO contacts_old (First_Name, Last_Name, Company, Business_Phone, Email_Address) VALUES ('Dave','Frank','Company1','0115 999999','zvv@zz.com'), ('Dave','Blogs','Company2','0115 888888','zvv@zz.com'), ('David','frank','Company3','0115 777777','zvv@zz.com')
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1
最佳答案
这只是一个框架，用于展示基于 JSON 中的数据构建查询的想法。这需要 根据您的实际数据和表架构进行一些微调和调整。
作为改进，我还会为每个表构建一个包含多个值的 INSERT。
编辑
更改代码以读取和插入 JSON 文件中的所有表。假定 JSON 文件中的表名和列名与实际表匹配。它还为具有 id 列的表添加了 ON DUPLICATE KEY UPDATE。
注意:未经测试，可能有拼写错误。
<?php
try
{
$connect = mysqli_connect("localhost", "fmart", "password", "mart_dbsync");
$query = '';
$table_data = '';
$filename = "single_table.json";
$data = file_get_contents($filename);
$array = json_decode($data, true);
foreach($array as $set)
{
$tblName = $set['tableName'];
if(sizeof($set['rows']) > 0) {
$query = '';
$colList = array();
$valList = array();
// Get list of column names
foreach($set['rows'][0] as $colname => $dataval) {
$colList[] = "`".$colName."`";
}
$query .= "INSERT INTO `".$tblName."` \n";
$query .= "(".implode(",",$colList).")\nVALUES\n";
// Go through the rows for this table.
foreach($set['rows'] as $idx => $row) {
$colDataA = array();
// Get the data values for this row.
foreach($row as $colName => $colData) {
$colDataA[] = "'".$colData."'";
}
$valList[] = "(".implode(",",$colDataA).")";
}
// Add values to the query.
$query .= implode(",\n",$valList)."\n";
// If id column present, add ON DUPLICATE KEY UPDATE clause
if(in_array("`id`",$colList)) {
$query .= "ON DUPLICATE KEY UPDATE\n\tSet ";
$tmp = array();
foreach($colList as $idx => $colName) {
$tmp[] = $colName." = new.".$colname." ";
}
$query .= implode(",",$tmp)."\n";
} else {
echo "<p><b>`id`</b> column not found. <i>ON DUPLICATE KEY UPDATE</i> clause <b>NOT</b> added.</p>\n";
echo "<p>Columns Found:<pre>".print_r($colList,true)."</pre></p>\n";
}
echo "<p>Insert query:<pre>$query</pre></p>";
$r = mysqli_query($connect, $query);
echo "<h1>".mysqli_num_rows($r)." Rows appeded in $tblName</h1>";
} else {
echo "<p>No rows to insert for $tblName</p>";
}
}
}
catch(Exception $e)
{
echo $e->getMessage();
}
?>
关于php - 如何使用php将多张表记录从json导入mysql，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/50942235/