根据solr文档，平局参数0应该使得分较低的字段对最终得分没有贡献：
当用户输入的术语针对多个字段进行测试时，可能会有多个字段匹配。如果是这样，每个字段将根据该单词在该字段中的常见程度(相对于所有其他文档，每个文档(生成不同的分数。通过平局参数，可以控制与得分最高的字段相比，得分较低的字段的得分对查询的最终得分的影响程度。
；0.0〃-默认值-使查询成为纯"；析取最大查询"：也就是说，只有得分最高的子查询才有助于最终得分。值"；1.0〃；使查询成为纯"；析取和查询"；其中最高得分的子查询是什么并不重要，因为最终得分将是子查询得分的总和。通常，低值(如0.1(是有用的。
然而，这对我不起作用。以下是我的搜索参数：
q:
parent_and_self_description:2-canal
parent_and_self_description:rct
info:2-canal
info:rct
qf: parent_and_self_description info^0.000000001
defType: edismax
尽管对信息的提升如此之低，但这里是最高结果的最终分数调试信息，由于其在"；信息"；字段：
9.731399 = sum of:
2.0064516 = weight(parent_and_self_description:canal in 753) [SchemaSimilarity], result of:
2.0064516 = score(freq=2.0), computed as boost * idf * tf from:
3.5589366 = idf, computed as log(1 + (N - n + 0.5) / (n + 0.5)) from:
75 = n, number of documents containing term
2651 = N, total number of documents with field
0.5637784 = tf, computed as freq / (freq + k1 * (1 - b + b * dl / avgdl)) from:
2.0 = freq, occurrences of term within document
1.2 = k1, term saturation parameter
0.75 = b, length normalization parameter
18.0 = dl, length of field
12.986043 = avgdl, average length of field
3.726406 = sum of:
1.7199546 = weight(parent_and_self_description:root in 753) [SchemaSimilarity], result of:
1.7199546 = score(freq=2.0), computed as boost * idf * tf from:
3.0507636 = idf, computed as log(1 + (N - n + 0.5) / (n + 0.5)) from:
125 = n, number of documents containing term
2651 = N, total number of documents with field
0.5637784 = tf, computed as freq / (freq + k1 * (1 - b + b * dl / avgdl)) from:
2.0 = freq, occurrences of term within document
1.2 = k1, term saturation parameter
0.75 = b, length normalization parameter
18.0 = dl, length of field
12.986043 = avgdl, average length of field
2.0064516 = weight(parent_and_self_description:canal in 753) [SchemaSimilarity], result of:
2.0064516 = score(freq=2.0), computed as boost * idf * tf from:
3.5589366 = idf, computed as log(1 + (N - n + 0.5) / (n + 0.5)) from:
75 = n, number of documents containing term
2651 = N, total number of documents with field
0.5637784 = tf, computed as freq / (freq + k1 * (1 - b + b * dl / avgdl)) from:
2.0 = freq, occurrences of term within document
1.2 = k1, term saturation parameter
0.75 = b, length normalization parameter
18.0 = dl, length of field
12.986043 = avgdl, average length of field
1.6317743 = weight(info:canal in 753) [SchemaSimilarity], result of:
1.6317743 = score(freq=3.0), computed as boost * idf * tf from:
3.226844 = idf, computed as log(1 + (N - n + 0.5) / (n + 0.5)) from:
2 = n, number of documents containing term
62 = N, total number of documents with field
0.50568736 = tf, computed as freq / (freq + k1 * (1 - b + b * dl / avgdl)) from:
3.0 = freq, occurrences of term within document
1.2 = k1, term saturation parameter
0.75 = b, length normalization parameter
56.0 = dl, length of field (approximate)
19.14516 = avgdl, average length of field
2.366767 = sum of:
0.7349925 = weight(info:root in 753) [SchemaSimilarity], result of:
0.7349925 = score(freq=1.0), computed as boost * idf * tf from:
2.8903718 = idf, computed as log(1 + (N - n + 0.5) / (n + 0.5)) from:
3 = n, number of documents containing term
62 = N, total number of documents with field
0.25428993 = tf, computed as freq / (freq + k1 * (1 - b + b * dl / avgdl)) from:
1.0 = freq, occurrences of term within document
1.2 = k1, term saturation parameter
0.75 = b, length normalization parameter
56.0 = dl, length of field (approximate)
19.14516 = avgdl, average length of field
1.6317743 = weight(info:canal in 753) [SchemaSimilarity], result of:
1.6317743 = score(freq=3.0), computed as boost * idf * tf from:
3.226844 = idf, computed as log(1 + (N - n + 0.5) / (n + 0.5)) from:
2 = n, number of documents containing term
62 = N, total number of documents with field
0.50568736 = tf, computed as freq / (freq + k1 * (1 - b + b * dl / avgdl)) from:
3.0 = freq, occurrences of term within document
1.2 = k1, term saturation parameter
0.75 = b, length normalization parameter
56.0 = dl, length of field (approximate)
19.14516 = avgdl, average length of field
我希望在两个字段上匹配的文档仅在parent_and_self_description字段上得分，并且我希望仅在信息字段上匹配文档的得分如此之低，以至于它们位于排名的最后。我做错了什么？非常感谢。
我只需要不在q参数中指定任何字段。所以不是
parent_and_self_description:2-canal
parent_and_self_description:rct
info:2-canal
info:rct
只做q=2-canal rct