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我正在使用 SparkSQL 加载一堆 JSON 文件，但有些有问题。

我想继续处理其他文件，同时忽略坏文件，我该怎么做？

我尝试使用 try-catch 但它仍然失败。例子：

try {
    val sqlContext = new org.apache.spark.sql.SQLContext(sc)
    import sqlContext._

    val jsonFiles=sqlContext.jsonFile("/requests.loading")
} catch {
    case _: Throwable => // Catching all exceptions and not doing anything with them
}

我失败了：

14/11/20 01:20:44 INFO scheduler.TaskSetManager: Starting task 3065.0 in stage 1.0 (TID 6150, HDdata2, NODE_LOCAL, 1246 bytes)<BR>
14/11/20 01:20:44 WARN scheduler.TaskSetManager: Lost task 3027.1 in stage 1.0 (TID 6130, HDdata2): com.fasterxml.jackson.core.JsonParseException: Unexpected end-of-input: was expecting closing quote for a string value
 at [Source: java.io.StringReader@753ab9f1; line: 1, column: 1805]
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1 回答 1

1

如果您使用的是 Spark 1.2，Spark SQL 将为您处理那些损坏的 JSON 记录。这是一个例子......

// requests.loading has some broken records
val jsonFiles=sqlContext.jsonFile("/requests.loading")
// Look at the schema of jsonFiles, you will see a new column called "_corrupt_record", which holds all broken JSON records
// jsonFiles.printSchema
// Register jsonFiles as a table
jsonFiles.registerTempTable("jsonTable")
// To query all normal records
sqlContext.sql("SELECT * FROM jsonTable WHERE _corrupt_record IS NULL")
// To query all broken JSON records
sqlContext.sql("SELECT _corrupt_record FROM jsonTable WHERE _corrupt_record IS NOT NULL")
于 2015-02-22T19:14:00.477 回答